Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me

Maybe we don't know that we should create a term that can be squared. But we do know that it would be easy to solve if we eliminate the $x$ term. We can use substitution of variables to eliminate it.
$ax^2+bx+c=0$
$x=y+d$
$a(y^2+2dy+d^2)+b(y+d)+c=0$
$ay^2+ad^2+(2ad+b)y+bd+c=0$
We want: $2ad+b=0$
$d=-\frac{b}{2a}$
$x=y-\frac{b}{2a}$
Substitute for $x$ in the original:
$a(y-\frac{b}{2a})^2+b(y-\frac{b}{2a})+c=0$
$ay^2-by+\frac{b^2}{4a}+by-\frac{b^2}{2a}+c=0$
$ay^2-\frac{b^2}{4a}+\frac{4ac}{4a}=0$
$y^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$
$y=\pm\frac{\sqrt{b^2-4ac}}{2a}$
Substitute for $y$ in $x=y-\frac{b}{2a}$:
$x=\pm\frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$
$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
