#### Answer

$y=\left\{ 7-2\sqrt{3},7+2\sqrt{3} \right\}$

#### Work Step by Step

The left side of the given equation, $
y^2-14y+49=12
,$ is a perfect square trinomial whose factored form is
\begin{array}{l}\require{cancel}
(y-7)^2=12
.\end{array}
Taking the square root of both sides (the Square Root Property), the solutions to the given equation are
\begin{array}{l}\require{cancel}
y-7=\pm\sqrt{12}
\\\\
y-7=\pm\sqrt{4\cdot3}
\\\\
y-7=\pm\sqrt{(2)^2\cdot3}
\\\\
y-7=\pm2\sqrt{3}
\\\\
y=7\pm2\sqrt{3}
.\end{array}
Hence, $
y=\left\{ 7-2\sqrt{3},7+2\sqrt{3} \right\}
.$