Introductory Algebra for College Students (7th Edition)

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