## Algebra 2 Common Core

$x=-\dfrac{3}{2}-i\dfrac{\sqrt{91}}{2}$ OR $x=-\dfrac{3}{2}+ i\dfrac{\sqrt{91}}{2}$
Adding $\left( \dfrac{b}{2} \right)^2$ on both sides of the given equation, $x^2+3x=-25 ,$ to complete the square of the left side results to \begin{align*} x^2+3x+\left( \dfrac{3}{2} \right)^2&=-25+\left( \dfrac{3}{2} \right)^2 \\\\ x^2+3x+\dfrac{9}{4}&=-25+\dfrac{9}{4} \\\\ \left( x+\dfrac{3}{2} \right)^2&=-\dfrac{100}{4}+\dfrac{9}{4} \\\\ \left( x+\dfrac{3}{2} \right)^2&=-\dfrac{91}{4} .\end{align*} Taking the square root of both sides (Square Root Property) and then solving for the variable, the equation above is equivalent to \begin{align*} x+\dfrac{3}{2}&=\pm\sqrt{-\dfrac{91}{4}} \\\\ x+\dfrac{3}{2}&=\pm i\sqrt{\dfrac{91}{4}} \\\\ x+\dfrac{3}{2}&=\pm i\dfrac{\sqrt{91}}{2} \\\\ x&=-\dfrac{3}{2}\pm i\dfrac{\sqrt{91}}{2} .\end{align*} Hence, the solutions are $x=-\dfrac{3}{2}-i\dfrac{\sqrt{91}}{2}$ OR $x=-\dfrac{3}{2}+ i\dfrac{\sqrt{91}}{2}$.