Solve by Factoring 4x^2=20x

$4 x^{2} = 20 x$

Step 1

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

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

Step 2

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

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

$4 x (x) - 20 x = 0$

Factor $4 x$ out of $- 20 x$ .

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

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

$4 x (x - 5) = 0$

Step 3

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

$x = 0$

$x - 5 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 6

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

$x = 0, 5$