To find the solution of a quadratic equation ax^{2 }+ bx + c = 0 we use the quadratic formula, which has the following form:

**x = [-b ± √ (b ^{2 }- 4ac)] / 2a**

Substituting the values of the coefficients a, b and c into it, we can easily obtain the values of x, remembering that «±» expresses that the equation has!**TWO SOLUTIONS**! Part "b^{2 }- 4ac ”is called discriminant and:

- if it is positive, there are TWO solutions.
- if it is zero there is only ONE solution.
- if it is negative there are two solutions that include imaginary numbers.

**Examples:**

**1. x² - 2x - 3 = 0**

The coefficients are: a = 1; b = - 2 and c = - 3. We check the value of the discriminant:

b^{2 }- 4ac = (- 2)^{2 }– 4(1)(- 3) = 4 + 12 = 16

Since 16> 0, there will be two different roots:

Now, substituting in the quadratic formula we have:

x = [- b ± √ (b^{2 }- 4ac)] / 2a → = {2 ± √16} / 2 = {2 ± 4} / 2

Therefore the two roots are:

x_{1 }= {2 + 4} / 2; x_{2 }= {2 – 4} / 2

x_{1 }= 3; x_{2 }= – 1

**2. x**^{2 }- 12x + 36 = 0.

^{2 }- 12x + 36 = 0.

The coefficients are: a = 1; b = - 12 and c = 36. We check the value of the discriminant:

b^{2 }- 4ac = (-12)^{2}- 4 (1) (36) = 144 - 144 = 0

Since the discriminant is equal to zero, there will be two roots of the same value:

Now, substituting in the quadratic formula we have:

x = (12 ± √0) / 2 = 12/2 = 6

x_{1 }= x_{2 }= 6

**3. x**^{2 }- 6x + 25 = 0.

^{2 }- 6x + 25 = 0.

The coefficients are: a = 1; b = - 6 and c = 25. We check the value of the discriminant:

b^{2 }- 4ac = (-6)^{2 }- 4 (1) (25) = 36 - 100 = - 64

Since 64 <0, there will be two imaginary roots:

Now, substituting in the quadratic formula we have:

x = [6 ± √ - 64] / 2

Remembering that i = √ -1, we have:

x = [6 ± (√ 64 · √ -1)] / 2 = (6 ± 8i) / 2

x_{1} = (6 + 8i) / 2 = 3 + 4i

x_{2} = (6 - 8i) / 2 = 3 - 4i