\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Everything SpawnofAngel says is right except that it is not y equals but x equals.
.
-b + or - sqrt(b^2-4(a)(c))
y= --------------------------------
2a
a= leading coeffiecent
b= second coeffiecent
c= last coeffiecent
such that y=ax2+bx+c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Everything SpawnofAngel says is right except that it is not y equals but x equals.