Solve 3x(x-1)=x(x+5)

$$3 x ( x - 1 ) = x ( x + 5 )$$

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 3x by x-1.

$$3x^{2}-3x=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$3x^{2}-3x=x^{2}+5x$$

Subtract x^{2} from both sides.

$$3x^{2}-3x-x^{2}=5x$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-3x=5x$$

Subtract 5x from both sides.

$$2x^{2}-3x-5x=0$$

Combine -3x and -5x to get -8x.

$$2x^{2}-8x=0$$

Factor out x.

$$x\left(2x-8\right)=0$$

To find equation solutions, solve x=0 and 2x-8=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x-1.

$$3x^{2}-3x=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$3x^{2}-3x=x^{2}+5x$$

Subtract x^{2} from both sides.

$$3x^{2}-3x-x^{2}=5x$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-3x=5x$$

Subtract 5x from both sides.

$$2x^{2}-3x-5x=0$$

Combine -3x and -5x to get -8x.

$$2x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 2}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±8}{4}$$

Now solve the equation x=\frac{8±8}{4} when ± is plus. Add 8 to 8.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{8±8}{4} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x-1.

$$3x^{2}-3x=x\left(x+5\right)$$

Use the distributive property to multiply x by x+5.

$$3x^{2}-3x=x^{2}+5x$$

Subtract x^{2} from both sides.

$$3x^{2}-3x-x^{2}=5x$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-3x=5x$$

Subtract 5x from both sides.

$$2x^{2}-3x-5x=0$$

Combine -3x and -5x to get -8x.

$$2x^{2}-8x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{0}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{0}{2}$$

Divide 0 by 2.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=4$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$