Solve for t 5t^2-20t+20=0

$5 t^{2} - 20 t + 20 = 0$

Step 1

Factor the left side of the equation.

Factor $5$ out of $5 t^{2} - 20 t + 20$ .

Factor $5$ out of $5 t^{2}$ .

$5 (t^{2}) - 20 t + 20 = 0$

Factor $5$ out of $- 20 t$ .

$5 (t^{2}) + 5 (- 4 t) + 20 = 0$

Factor $5$ out of $20$ .

$5 t^{2} + 5 (- 4 t) + 5 \cdot 4 = 0$

Factor $5$ out of $5 t^{2} + 5 (- 4 t)$ .

$5 (t^{2} - 4 t) + 5 \cdot 4 = 0$

Factor $5$ out of $5 (t^{2} - 4 t) + 5 \cdot 4$ .

$5 (t^{2} - 4 t + 4) = 0$

Factor using the perfect square rule.

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

$5 (t^{2} - 4 t + 2^{2}) = 0$

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

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

Rewrite the polynomial.

$5 (t^{2} - 2 \cdot t \cdot 2 + 2^{2}) = 0$

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

$5 {(t - 2)}^{2} = 0$

Step 2

Divide each term in $5 {(t - 2)}^{2} = 0$ by $5$ and simplify.

Divide each term in $5 {(t - 2)}^{2} = 0$ by $5$ .

$\frac{5 {(t - 2)}^{2}}{5} = \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 {(t - 2)}^{2}}{5} = \frac{0}{5}$

Divide ${(t - 2)}^{2}$ by $1$ .

${(t - 2)}^{2} = \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

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

Step 3

Set the $t - 2$ equal to $0$ .

$t - 2 = 0$

Step 4

Add $2$ to both sides of the equation.

$t = 2$