Solve for t t^2-6t+9=0

$t^{2} - 6 t + 9 = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$t^{2} - 6 t + 3^{2} = 0$

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

$6 t = 2 \cdot t \cdot 3$

Rewrite the polynomial.

$t^{2} - 2 \cdot t \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = t$ and $b = 3$ .

${(t - 3)}^{2} = 0$

Set the $t - 3$ equal to $0$ .

$t - 3 = 0$

Add $3$ to both sides of the equation.

$t = 3$