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