# Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 596: 110

$x=-\dfrac{b}{2} \pm \dfrac{\sqrt{b^2-4c}}{2}$

#### 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, $x=-\dfrac{b}{2} \pm \dfrac{\sqrt{b^2-4c}}{2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.