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In mathematics, a polynomial equation with second degree represents a quadratic equation. Quadratic equation in its standard form is given by,
$Ax^2+Bx+C=0$
where, A, B and C are constant, x is variable and A$\neq$0.

Solving quadratic equation gives you two solutions, they are known as roots of the equation. These roots may be imaginary or real. These roots can be determined by finding the value of discriminant, given by the formula,
D = $b^2$ - 4ac

The quadratic formula calculator uses the quadratic formula and give's you the roots of the given equation.

## Steps to Find the Quadratic Formula

Step 1: Note down the constants A, B and C.

Step 2: Using the formula D = $b^2$ - 4ac, find the discriminant value.
If D<0, then the roots are imaginary and if D$\geq$0, it is real.

Step 3: Find the roots of the equation by the following equation,
$x_1$=$\frac{-b+\sqrt{D}}{2a}$ and $x_2$=$\frac{-b-\sqrt{D}}{2a}$

Given below are some of the examples based on quadratic formula:

### Solved Examples

Question 1: Solve the equation $x^2+2x-3=0$.

Solution:

Step 1: Given equation : $x^2+2x-3=0$
A = 1, B = 2 and C = -3

Step 2: D = $b^2$ - 4ac
D = $(2)^2$ - 4(1)(-3)
D = 4 + 12
D = 16

Step 3: As D>0, roots are real.
The roots of the given equation are,

$x_1$=$\frac{-b+\sqrt{D}}{2a}$

$x_1$=$\frac{-2+\sqrt{16}}{2(1)}$

$x_1$=$\frac{-2+4}{2}$

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

$x_1=1$

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

$x_1$=$\frac{-2-\sqrt{16}}{2(1)}$

$x_1$=$\frac{-2-4}{2}$

$x_1$=$\frac{-6}{2}$

$x_1=-3$

Solution : $x_1=1$ and $x_1=-3$

Question 2: Solve the equation $3x^2-10x+5=0$.

Solution:

Step 1: Given equation : $3x^2-10x+5=0$
A = 3, B = -10 and C = 5

Step 2: D = $b^2$ - 4ac
D = $(-10)^2$ - 4(3)(5)
D = 100 - 60
D = 40

Step 3: As D>0, roots are real.
The roots of the given equation are,

$x_1$=$\frac{-b+\sqrt{D}}{2a}$

$x_1$=$\frac{-(-10)+\sqrt{40}}{2(3)}$

$x_1$=$\frac{10+2\sqrt{10}}{6}$

$x_1$=$\frac{5+\sqrt{10}}{3}$

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

$x_1$=$\frac{-(-10)-\sqrt{40}}{2(3)}$

$x_1$=$\frac{10-2\sqrt{10}}{6}$

$x_1$=$\frac{5-\sqrt{10}}{3}$

Solution : $x_1$=$\frac{5+\sqrt{10}}{3}$ and $x_1$=$\frac{5-\sqrt{10}}{3}$