## Algebra 1

$x = (-b±\sqrt{(b^2-4ac)})/2a$
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) $x^2+b/a*x = -c/a$ $x^2+b/a*x+(b/2a)^2 = -c/a+(b/2a)^2$ $x^2+b/a*x+b^2/4a^2 = -c/a+b^2/4a^2$ $(x+b/2a)^2 = b^2/4a^2-c/a$ $(x+b/2a)^2 = b^2/4a^2-c/a*(4a/4a)$ $(x+b/2a)^2 = b^2/4a^2-4ac/4a^2$ $(x+b/2a)^2 = (b^2-4ac)/4a^2$ $\sqrt {(x+b/2a)^2} = \sqrt{(b^2-4ac)/4a^2}$ $x+b/2a = ±\sqrt{(b^2-4ac)/4a^2}$ $x+b/2a = ±\sqrt{(b^2-4ac)}/2a$ $x+b/2a-b/2a = ±\sqrt{(b^2-4ac)}/2a-b/2a$ $x = (-b±\sqrt{(b^2-4ac)})/2a$