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Quadratic equation is a polynomial equation with second degree, is in the form of $Ax^2 + Bx + C = 0$. A, B and C are the known values and x is the unknown variable. Quadratic equation finds the factors of the given equation, they are known as roots of the equation.

This online Quadratic equation calculator calculate the discriminant and finds the roots of the given equation.

## Steps to Find the Quadratic Equation

Step 1: Note down the values of A,B and C from the given problem.

Step 2: Determine the discriminant D by using the formula,
$D = b^2 - 4ac$

If the D < 0, then the roots does not exists. If D $\geq$ 0, then roots exists.

Step 3: To find the roots of the equation, we use the following formula,
$x$ = $\frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

Below are the problems based on Quadratic Equation Calculator.

### Solved Examples

Question 1: Determine the roots of the equation 2x^2 + 7x - 15 = 0$Solution: Step 1: A = 2, B = 7, c = -15 Step 2:$D = b^2 - 4acD = 7^2 - 4(2)(-15)D = 49 + 169 D = 169$D > 0 i.e., roots exists. The roots of the equation are,$x_1$=$\frac{-b + \sqrt{D}}{2a}x_1$=$\frac{-7 + \sqrt{169}}{2(2)}x_1$=$\frac{-7 + 13}{4}x_1$=$\frac{6}{4}x_1$=$\frac{3}{2}x_2$=$\frac{-b - \sqrt{D}}{2a}x_2$=$\frac{-7 - \sqrt{169}}{2(2)}x_2$=$\frac{-7 - 13}{4}x_2$=$\frac{-20}{4}x_2 = -5$Therefore, the roots of$2x^2 + 7x - 15 = 0$are ($\frac{3}{2}$, -5). Question 2: Determine the roots of the equation 5x^2 + 6x + 1 = 0$
Solution:

Step 1: A = 5, B = 6, c = 1

Step 2: $D = b^2 - 4ac$

$D = 6^2 - 4(5)(1)$

$D = 36 - 20$

$D = 16$

D > 0 i.e., roots exists.

The roots of the equation are,
$x_1$ = $\frac{-b + \sqrt{D}}{2a}$

$x_1$ = $\frac{-6 + \sqrt{16}}{2(5)}$

$x_1$ =$\frac{-6 + 4}{10}$

$x_1$ = $\frac{-2}{10}$

$x_1$ = $\frac{-1}{5}$

$x_2$ = $\frac{-b - \sqrt{D}}{2a}$

$x_2$ = $\frac{-6 - \sqrt{16}}{2(5)}$

$x_2$ = $\frac{-6 - 4}{10}$

$x_2$ = $\frac{-10}{10}$

$x_2 = -1$

Therefore, the roots of $2x^2 + 7x - 15 = 0$ are ($\frac{-1}{5}$, -1).