Find the Roots (Zeros) y=x^2+12x+36

$y = x^{2} + 12 x + 36$

Set $x^{2} + 12 x + 36$ equal to $0$ .

$x^{2} + 12 x + 36 = 0$

Solve for $x$ .

Factor using the perfect square rule.

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

$x^{2} + 12 x + 6^{2} = 0$

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

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 6 + 6^{2} = 0$

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

${(x + 6)}^{2} = 0$

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$