Start with the equation
[tex]ax^2+bx+c=0[/tex]
Factor [tex]a[/tex] from the first two terms:
[tex]a\left(x^2+\dfrac bax\right)+c=0[/tex]
Complete the square; do this by adding an appropriate constant to [tex]x^2+\dfrac bax[/tex] to form a perfect square trinomial. Recall that
[tex](s+t)^2=s^2+2st+t^2[/tex]
So here we have
[tex]\begin{cases}s=x\\\\2st=\dfrac ba x\end{cases}\implies t=\dfrac b{2a}[/tex]
and we have to add [tex]t^2=\dfrac{b^2}{4a^2}[/tex] to make the perfect square. But we also have to subtract this same constant to preserve equality:
[tex]a\left(x^2+\dfrac bax+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}\right)+c=0[/tex]
[tex]a\left(\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2}{4a^2}\right)+c=0[/tex]
[tex]a\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2}{4a}+c=0[/tex]
[tex]\left(x+\dfrac b{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}[/tex]
Now we can take the square root and solve for [tex]x[/tex]:
[tex]x+\dfrac b{2a}=\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}[/tex]
[tex]x=-\dfrac b{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}[/tex]
[tex]x=-\dfrac b{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}[/tex]
You should recognize this formula...
