Answer
$t=\left\{ \dfrac{-2-2\sqrt{3}}{5},\dfrac{-2+2\sqrt{3}}{5} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$