Solve by Factoring 5x^2-20x+20=0

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

Step 1

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

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

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

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

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

Factor $5$ out of $20$ .

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

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

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

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

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

Step 2

Factor using the perfect square rule.

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

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

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

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

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

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

Step 3

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

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

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $0$ by $5$ .

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

Step 4

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

$x - 2 = 0$

Step 5

Add $2$ to both sides of the equation.

$x = 2$