### Theory:

A quadratic equation in the variable $$x$$ is an equation of the form $$ax^2 + bx + c = 0$$, where $$a$$, $$b$$ and $$c$$ are numbers, $$a \ne 0$$. The degree of the quadratic equation is $$2$$.
Important!
The equation $$ax^2 + bx + c = 0$$ is called the standard form of a quadratic equation.
The value of $$x$$ that makes the expression $$ax^2 + bx + c$$ is zero, called the roots of the quadratic equation.
Consider the quadratic equation $$ax^2 + bx + c = 0$$, where $$a \ne 0$$

Divide the equation by $$a$$.

${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$

Move the constant to the right side.

${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$

Add the square of one half of the coefficient of $$x$$ on both sides.

${x}^{2}+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$

${\left(x+\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+\frac{{b}^{2}}{4{a}^{2}}$

${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4\mathit{ac}}{4{a}^{2}}$

Taking square root on both sides.

$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}-4\mathit{ac}}{4{a}^{2}}}$

$x+\frac{b}{2a}=±\frac{\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$

$x=-\frac{b}{2a}±\frac{\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$

$x=\frac{-b±\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$

Therefore, the roots of $$ax^2 + bx + c = 0$$ are $x=\frac{-b+\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$ and $x=\frac{-b-\sqrt{{b}^{2}-4\mathit{ac}}}{2a}$.