Precalculus (6th Edition)

Option $(B)$ $\color{blue}{x=\left\{\frac{-5+\sqrt7}{2}, \frac{-5-\sqrt7}{2}\right\}}$
RECALL: If $x^2=a$, then $x = \pm \sqrt{a}$ Among the given equation, the one in Option (B) is set up for direct use of the square root property. Using the property gives: $\sqrt{(2x+5)^2}=\pm \sqrt{7} \\2x+5 = \pm \sqrt{7}$ Subtract $5$ to both sides: $2x=-5 \pm \sqrt{7}$ Divide both sides by $2$: $x = \dfrac{-5\pm\sqrt{7}}{2}$ Thus, $\color{blue}{x=\left\{\frac{-5+\sqrt7}{2}, \frac{-5-\sqrt7}{2}\right\}}$.