Solve for y y^2=12y-36

$y^{2} = 12 y - 36$

Subtract $12 y$ from both sides of the equation.

$y^{2} - 12 y = - 36$

Add $36$ to both sides of the equation.

$y^{2} - 12 y + 36 = 0$

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$y^{2} - 12 y + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 y = 2 \cdot y \cdot 6$

Rewrite the polynomial.

$y^{2} - 2 \cdot y \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 6$ .

${(y - 6)}^{2} = 0$

Set the $y - 6$ equal to $0$ .

$y - 6 = 0$

Add $6$ to both sides of the equation.

$y = 6$