Solve for x 20x=5x^2+20

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

Subtract $5 x^{2}$ from both sides of the equation.

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

Subtract $20$ from both sides of the equation.

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

Factor the left side of the equation.

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

Reorder $20 x$ and $- 5 x^{2}$ .

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

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 $- 5 (x^{2} - 4 x) - 5 (4)$ .

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

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 $b = 2$ .

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

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 $- 5$ .

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$