Solve for x x^2+20x+100=0

$x^{2} + 20 x + 100 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $100$ as $10^{2}$ .

$x^{2} + 20 x + 10^{2} = 0$

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

$20 x = 2 \cdot x \cdot 10$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 10 + 10^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 10$ .

${(x + 10)}^{2} = 0$

Step 2

Set the $x + 10$ equal to $0$ .

$x + 10 = 0$

Step 3

Subtract $10$ from both sides of the equation.

$x = - 10$