## Intermediate Algebra (12th Edition)

$t=\left\{ \dfrac{-2-2\sqrt{3}}{5},\dfrac{-2+2\sqrt{3}}{5} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(5t+2)^2=12 ,$ take the square root of both sides (Square Root Property) and simplify the radical. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} 5t+2=\pm\sqrt{12} .\end{array} Simplifying the radical and then using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 5t+2=\pm\sqrt{4\cdot3} \\\\ 5t+2=\pm\sqrt{(2)^2\cdot3} \\\\ 5t+2=\pm2\sqrt{3} \\\\ 5t=-2\pm2\sqrt{3} \\\\ t=\dfrac{-2\pm2\sqrt{3}}{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=\dfrac{-2-2\sqrt{3}}{5} \\\\\text{OR}\\\\ t=\dfrac{-2+2\sqrt{3}}{5} .\end{array} Hence, $t=\left\{ \dfrac{-2-2\sqrt{3}}{5},\dfrac{-2+2\sqrt{3}}{5} \right\} .$