Solve by Factoring 20x^2=4x

$20 x^{2} = 4 x$

Step 1

Subtract $4 x$ from both sides of the equation.

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

Step 2

Factor $4 x$ out of $20 x^{2} - 4 x$ .

Factor $4 x$ out of $20 x^{2}$ .

$4 x (5 x) - 4 x = 0$

Factor $4 x$ out of $- 4 x$ .

$4 x (5 x) + 4 x (- 1) = 0$

Factor $4 x$ out of $4 x (5 x) + 4 x (- 1)$ .

$4 x (5 x - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$5 x - 1 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $5 x - 1$ equal to $0$ and solve for $x$ .

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Solve $5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 6

The final solution is all the values that make $4 x (5 x - 1) = 0$ true.

$x = 0, \frac{1}{5}$