Answer
$x=-\dfrac{b}{2} \pm \dfrac{\sqrt{b^2-4c}}{2}$
Work Step by Step
Need to solve for $y$.
We have $x^2+bx+c=0$ or, $x^2+bx=-c$
To complete the square we will have to add $(\dfrac{b}{2})^2$.
Thus, $x^2+x+(\dfrac{b}{2})^2=(\dfrac{b}{2})^2-c$
or, $x^2+2(x)(\dfrac{1}{2})+\dfrac{b}{4}=\dfrac{b^2}{4}-c$
or, $(x+\dfrac{b}{2})^2=\dfrac{b^2-4c}{4}$
or, $(x+\dfrac{b}{2})=\pm \sqrt{\dfrac{b^2-4c}{4}}$
Hence, $x=-\dfrac{b}{2} \pm \dfrac{\sqrt{b^2-4c}}{2}$
