Answer
$x=\frac{-b\pm \sqrt b^2-4c}{2}$
Work Step by Step
Given: $x^2+bx+c=0$
$(x^2+bx)+c=0$
$(x^2+2\times\frac{b}{2}\times x+(\frac{b}{2})^2)-(\frac{b}{2})^2+c=0$
$(x+\frac{b}{2})^2+\frac{4c-b^2}{4}=0$
$(x+\frac{b}{2})^2=\frac{b^2-4c}{4}$
$x+\frac{b}{2}=\pm\sqrt \frac{b^2-4c}{4}$
$x+\frac{b}{2}=\pm \frac{\sqrt b^2-4c}{2}$
$x=-\frac{b}{2}\pm \frac{\sqrt b^2-4c}{2}$
$x=\frac{-b\pm \sqrt {b^2-4c}}{2}$
