How can you solve any quadratic equation?

The most general methods which will cope with any quadratic equation in one variable are:
The quadratic formula.
Completing the square.
Whether the roots are integers, rational, irrational, or even non-Real Complex numbers, both methods can both locate them.
Quadratic formula
The roots of ${\displaystyle a{x}^2+bx+c=0}$ are given by the formula:
${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
Completing the square
Given:
${\displaystyle a{x}^2+bx+c=0}$
Note that:
${\displaystyle a{(x+\frac{b}{2a})}^2=a{x}^2+bx+\frac{b^2}{4a}}$
We can now rewrite our equation as follows:
${\displaystyle a{(x+\frac{b}{2a})}^2=\frac{b^2}{4a}-c}$
Dividing both sides by a we find:
${\displaystyle {(x+\frac{b}{2a})}^2=\frac{b^2}{4a^2}-\frac{c}{a}}$
Thus:
${\displaystyle x+\frac{b}{2a}=\pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}}$
Which can be simplified to:
${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
So the quadratic formula and completing the square are somewhat equivalent, however in some situations completing the square can be a little cleaner:
For example, factoring ${\displaystyle x^2+4x-21}$ by completing the square:
${\displaystyle x^2+4x-21}$
${\displaystyle =x^2+4x+4-25}$
${\displaystyle ={(x+2)}^2-5^2}$
${\displaystyle =((x+2)-5)((x+2)+5)}$
${\displaystyle =(x-3)(x+7)}$
You may also use the quadratic formula:
${\displaystyle x^2+4x-21}$ is ${\displaystyle a{x}^2+bx+c}$ with ${\displaystyle a=1}$, ${\displaystyle b=4}$, ${\displaystyle c=-21}$
thus has zeros:
${\displaystyle x=\frac{-4\pm \sqrt{4^2-(4\cdot 1\cdot (-21))}}{2\cdot 1}}$
${\displaystyle =\frac{-4\pm \sqrt{16+84}}{2}}$
${\displaystyle =\frac{-4\pm \sqrt{100}}{2}}$
${\displaystyle =\frac{-4\pm 10}{2}}$
${\displaystyle =-2\pm 5}$
i.e. ${\displaystyle x=-7}$ and ${\displaystyle x=3}$
Therefore factors:
${\displaystyle (x+7)(x-3)}$