# X2 5x 3 Solve for x (complex solution)
x=\frac{5+\sqrt{11}i}{6}\approx 0.833333333+0.552770798i
x=\frac{-\sqrt{11}i+5}{6}\approx 0.833333333-0.552770798i All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3\times 3}}{2\times 3}
x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\times 3}}{2\times 3}
x=\frac{-\left(-5\right)±\sqrt{25-12\times 3}}{2\times 3}
x=\frac{-\left(-5\right)±\sqrt{25-36}}{2\times 3}
x=\frac{-\left(-5\right)±\sqrt{-11}}{2\times 3}
Take the square root of -11.
x=\frac{-\left(-5\right)±\sqrt{11}i}{2\times 3}
x=\frac{5±\sqrt{11}i}{2\times 3}
Now solve the equation x=\frac{5±\sqrt{11}i}{6} when ± is plus. Add 5 to i\sqrt{11}.
Now solve the equation x=\frac{5±\sqrt{11}i}{6} when ± is minus. Subtract i\sqrt{11} from 5.
x=\frac{-\sqrt{11}i+5}{6}
The equation is now solved.
x=\frac{5+\sqrt{11}i}{6} x=\frac{-\sqrt{11}i+5}{6}

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y=x^{2}-5x 3

en

Sours: https://www.symbolab.com/solver/equation-calculator/y=x%5E%7B2%7D-5x%203

### Step  2  :

#### Pulling out like terms :

2.1     Pull out like factors :

-x2 - 5x - 3  =   -1 • (x2 + 5x + 3)

#### Trying to factor by splitting the middle term

2.2     Factoring  x2 + 5x + 3

The first term is,  x2  its coefficient is  1 .
The middle term is,  +5x  its coefficient is  5 .
The last term, "the constant", is  +3

Step-1 : Multiply the coefficient of the first term by the constant   1 • 3 = 3

Step-2 : Find two factors of  3  whose sum equals the coefficient of the middle term, which is   5 .

 -3 + -1 = -4 -1 + -3 = -4 1 + 3 = 4 3 + 1 = 4

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

-x2 - 5x - 3 = 0

### Step  3  :

#### Parabola, Finding the Vertex :

3.1      Find the Vertex of   y = -x2-5x-3

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens down and accordingly has a highest point (AKA absolute maximum) .    We know this even before plotting  "y"  because the coefficient of the first term, -1 , is negative (smaller than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -2.5000

Plugging into the parabola formula  -2.5000  for  x  we can calculate the  y -coordinate :
y = -1.0 * -2.50 * -2.50 - 5.0 * -2.50 - 3.0
or   y = 3.250

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = -x2-5x-3
Axis of Symmetry (dashed)  {x}={-2.50}
Vertex at  {x,y} = {-2.50, 3.25}
x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.70, 0.00}
Root 2 at  {x,y} = {-4.30, 0.00}

#### Solve Quadratic Equation by Completing The Square

3.2     Solving   -x2-5x-3 = 0 by Completing The Square .

Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
x2+5x+3 = 0  Subtract  3  from both side of the equation :
x2+5x = -3

Now the clever bit: Take the coefficient of  x , which is  5 , divide by two, giving  5/2 , and finally square it giving  25/4

Add  25/4  to both sides of the equation :
On the right hand side we have :
-3  +  25/4    or,  (-3/1)+(25/4)
The common denominator of the two fractions is  4   Adding  (-12/4)+(25/4)  gives  13/4
So adding to both sides we finally get :
x2+5x+(25/4) = 13/4

Adding  25/4  has completed the left hand side into a perfect square :
x2+5x+(25/4)  =
(x+(5/2)) • (x+(5/2))  =
(x+(5/2))2
Things which are equal to the same thing are also equal to one another. Since
x2+5x+(25/4) = 13/4 and
x2+5x+(25/4) = (x+(5/2))2
then, according to the law of transitivity,
(x+(5/2))2 = 13/4

We'll refer to this Equation as  Eq. #3.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x+(5/2))2  is
(x+(5/2))2/2 =
(x+(5/2))1 =
x+(5/2)

Now, applying the Square Root Principle to  Eq. #3.2.1  we get:
x+(5/2) = √ 13/4

Subtract  5/2  from both sides to obtain:
x = -5/2 + √ 13/4

Since a square root has two values, one positive and the other negative
x2 + 5x + 3 = 0
has two solutions:
x = -5/2 + √ 13/4
or
x = -5/2 - √ 13/4

Note that  √ 13/4 can be written as
√ 13  / √ 4   which is √ 13  / 2

3.3     Solving    -x2-5x-3 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

- B  ±  √ B2-4AC
x =   ————————
2A

In our case,  A   =     -1
B   =    -5
C   =   -3

Accordingly,  B2  -  4AC   =
25 - 12 =
13

5 ± √ 13
x  =    —————
-2

√ 13   , rounded to 4 decimal digits, is   3.6056
So now we are looking at:
x  =  ( 5 ±  3.606 ) / -2

Two real solutions:

x =(5+√13)/-2=-4.303

or:

x =(5-√13)/-2=-0.697

### Two solutions were found :

1.  x =(5-√13)/-2=-0.697
2.  x =(5+√13)/-2=-4.303
Sours: https://www.tiger-algebra.com/drill/-x~2-5x-3=0/
27 Rozwiąż nierówność 2x2−5x+3≤0

### Reformatting the input :

(1): "x2"   was replaced by   "x^2".

### Step  1  :

#### Trying to factor by splitting the middle term

1.1     Factoring  x2-5x-3

The first term is,  x2  its coefficient is  1 .
The middle term is,  -5x  its coefficient is  -5 .
The last term, "the constant", is  -3

Step-1 : Multiply the coefficient of the first term by the constant   1 • -3 = -3

Step-2 : Find two factors of  -3  whose sum equals the coefficient of the middle term, which is   -5 .

 -3 + 1 = -2 -1 + 3 = 2

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

x2 - 5x - 3 = 0

### Step  2  :

#### Parabola, Finding the Vertex :

2.1      Find the Vertex of   y = x2-5x-3

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   2.5000

Plugging into the parabola formula   2.5000  for  x  we can calculate the  y -coordinate :
y = 1.0 * 2.50 * 2.50 - 5.0 * 2.50 - 3.0
or   y = -9.250

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-5x-3
Axis of Symmetry (dashed)  {x}={ 2.50}
Vertex at  {x,y} = { 2.50,-9.25}
x -Intercepts (Roots) :
Root 1 at  {x,y} = {-0.54, 0.00}
Root 2 at  {x,y} = { 5.54, 0.00}

#### Solve Quadratic Equation by Completing The Square

2.2     Solving   x2-5x-3 = 0 by Completing The Square .

Add  3  to both side of the equation :
x2-5x = 3

Now the clever bit: Take the coefficient of  x , which is  5 , divide by two, giving  5/2 , and finally square it giving  25/4

Add  25/4  to both sides of the equation :
On the right hand side we have :
3  +  25/4    or,  (3/1)+(25/4)
The common denominator of the two fractions is  4   Adding  (12/4)+(25/4)  gives  37/4
So adding to both sides we finally get :
x2-5x+(25/4) = 37/4

Adding  25/4  has completed the left hand side into a perfect square :
x2-5x+(25/4)  =
(x-(5/2)) • (x-(5/2))  =
(x-(5/2))2
Things which are equal to the same thing are also equal to one another. Since
x2-5x+(25/4) = 37/4 and
x2-5x+(25/4) = (x-(5/2))2
then, according to the law of transitivity,
(x-(5/2))2 = 37/4

We'll refer to this Equation as  Eq. #2.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x-(5/2))2  is
(x-(5/2))2/2 =
(x-(5/2))1 =
x-(5/2)

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
x-(5/2) = √ 37/4

Add  5/2  to both sides to obtain:
x = 5/2 + √ 37/4

Since a square root has two values, one positive and the other negative
x2 - 5x - 3 = 0
has two solutions:
x = 5/2 + √ 37/4
or
x = 5/2 - √ 37/4

Note that  √ 37/4 can be written as
√ 37  / √ 4   which is √ 37  / 2

2.3     Solving    x2-5x-3 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

- B  ±  √ B2-4AC
x =   ————————
2A

In our case,  A   =     1
B   =    -5
C   =   -3

Accordingly,  B2  -  4AC   =
25 - (-12) =
37

5 ± √ 37
x  =    —————
2

√ 37   , rounded to 4 decimal digits, is   6.0828
So now we are looking at:
x  =  ( 5 ±  6.083 ) / 2

Two real solutions:

x =(5+√37)/2= 5.541

or:

x =(5-√37)/2=-0.541

### Two solutions were found :

1.  x =(5-√37)/2=-0.541
2.  x =(5+√37)/2= 5.541
Sours: https://www.tiger-algebra.com/drill/x2-5x-3=0/

## 3 x2 5x

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x^2+5x+3=0 by completing the square, neatly typed, 2.5 minutes

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