Answer
$x = (-b±\sqrt{(b^2-4ac)})/2a$
Work Step by Step
Given formula is $ax^2+bx+c=0$
$ax^2+bx+c=0$
$ax^2+bx+c-c=0-c$
$ax^2+bx=-c$
$(ax^2+bx)/a=-c/a$ (coefficient of $a$ is now 1)
$x^2+b/a*x = -c/a$
$x^2+b/a*x+(b/2a)^2 = -c/a+(b/2a)^2$
$x^2+b/a*x+b^2/4a^2 = -c/a+b^2/4a^2$
$(x+b/2a)^2 = b^2/4a^2-c/a$
$(x+b/2a)^2 = b^2/4a^2-c/a*(4a/4a)$
$(x+b/2a)^2 = b^2/4a^2-4ac/4a^2$
$(x+b/2a)^2 = (b^2-4ac)/4a^2$
$\sqrt {(x+b/2a)^2} = \sqrt{(b^2-4ac)/4a^2}$
$x+b/2a = ±\sqrt{(b^2-4ac)/4a^2}$
$x+b/2a = ±\sqrt{(b^2-4ac)}/2a$
$x+b/2a-b/2a = ±\sqrt{(b^2-4ac)}/2a-b/2a$
$x = (-b±\sqrt{(b^2-4ac)})/2a$
