Solve by Factoring 25x^2=20x-4

$25 x^{2} = 20 x - 4$

Move all the expressions to the left side of the equation.

Subtract $20 x$ from both sides of the equation.

$25 x^{2} - 20 x = - 4$

Add $4$ to both sides of the equation.

$25 x^{2} - 20 x + 4 = 0$

Factor using the perfect square rule.

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

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

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

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

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

$20 x = 2 \cdot (5 x) \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 5 x$ and $b = 2$ .

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

Set the $5 x - 2$ equal to $0$ .

$5 x - 2 = 0$

Solve for $x$ .

Add $2$ to both sides of the equation.

$5 x = 2$

Divide each term in $5 x = 2$ by $5$ and simplify.

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Divide $x$ by $1$ .

$x = \frac{2}{5}$