Solve 3xleft(x-4right)=4x(x-4)

$$3x \left( x-4 \right) = 4x(x-4)$$

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 3x by x-4.

$$3x^{2}-12x=4x\left(x-4\right)$$

Use the distributive property to multiply 4x by x-4.

$$3x^{2}-12x=4x^{2}-16x$$

Subtract 4x^{2} from both sides.

$$3x^{2}-12x-4x^{2}=-16x$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-12x=-16x$$

Add 16x to both sides.

$$-x^{2}-12x+16x=0$$

Combine -12x and 16x to get 4x.

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

Factor out x.

$$x\left(-x+4\right)=0$$

To find equation solutions, solve x=0 and -x+4=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by x-4.

$$3x^{2}-12x=4x\left(x-4\right)$$

Use the distributive property to multiply 4x by x-4.

$$3x^{2}-12x=4x^{2}-16x$$

Subtract 4x^{2} from both sides.

$$3x^{2}-12x-4x^{2}=-16x$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-12x=-16x$$

Add 16x to both sides.

$$-x^{2}-12x+16x=0$$

Combine -12x and 16x to get 4x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}}}{2\left(-1\right)}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±4}{-2}$$

Now solve the equation x=\frac{-4±4}{-2} when ± is plus. Add -4 to 4.

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

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-2} when ± is minus. Subtract 4 from -4.

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

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=0 x=4$$

Steps for Completing the Square

Use the distributive property to multiply 3x by x-4.

$$3x^{2}-12x=4x\left(x-4\right)$$

Use the distributive property to multiply 4x by x-4.

$$3x^{2}-12x=4x^{2}-16x$$

Subtract 4x^{2} from both sides.

$$3x^{2}-12x-4x^{2}=-16x$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-12x=-16x$$

Add 16x to both sides.

$$-x^{2}-12x+16x=0$$

Combine -12x and 16x to get 4x.

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

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=\frac{0}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=\frac{0}{-1}$$

Divide 0 by -1.

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

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.

$$x=4 x=0$$