Solve for t t^2-8t+16=0

$t^{2} - 8 t + 16 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$t^{2} - 8 t + 4^{2} = 0$

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

$8 t = 2 \cdot t \cdot 4$

Rewrite the polynomial.

$t^{2} - 2 \cdot t \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = t$ and $b = 4$ .

${(t - 4)}^{2} = 0$

Step 2

Set the $t - 4$ equal to $0$ .

$t - 4 = 0$

Step 3

Add $4$ to both sides of the equation.

$t = 4$