Solve (x+3)x=2x(6-x)

$$(x+3)x=2x(6-x)$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x=2x\left(6-x\right)$$

Use the distributive property to multiply 2x by 6-x.

$$x^{2}+3x=12x-2x^{2}$$

Subtract 12x from both sides.

$$x^{2}+3x-12x=-2x^{2}$$

Combine 3x and -12x to get -9x.

$$x^{2}-9x=-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-9x+2x^{2}=0$$

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

$$3x^{2}-9x=0$$

Factor out x.

$$x\left(3x-9\right)=0$$

To find equation solutions, solve x=0 and 3x-9=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x=2x\left(6-x\right)$$

Use the distributive property to multiply 2x by 6-x.

$$x^{2}+3x=12x-2x^{2}$$

Subtract 12x from both sides.

$$x^{2}+3x-12x=-2x^{2}$$

Combine 3x and -12x to get -9x.

$$x^{2}-9x=-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-9x+2x^{2}=0$$

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

$$3x^{2}-9x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}}}{2\times 3}$$

Take the square root of \left(-9\right)^{2}.

$$x=\frac{-\left(-9\right)±9}{2\times 3}$$

The opposite of -9 is 9.

$$x=\frac{9±9}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{9±9}{6}$$

Now solve the equation x=\frac{9±9}{6} when ± is plus. Add 9 to 9.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{9±9}{6} when ± is minus. Subtract 9 from 9.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=3 x=0$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by x.

$$x^{2}+3x=2x\left(6-x\right)$$

Use the distributive property to multiply 2x by 6-x.

$$x^{2}+3x=12x-2x^{2}$$

Subtract 12x from both sides.

$$x^{2}+3x-12x=-2x^{2}$$

Combine 3x and -12x to get -9x.

$$x^{2}-9x=-2x^{2}$$

Add 2x^{2} to both sides.

$$x^{2}-9x+2x^{2}=0$$

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

$$3x^{2}-9x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-9x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{9}{3}\right)x=\frac{0}{3}$$

Divide -9 by 3.

$$x^{2}-3x=\frac{0}{3}$$

Divide 0 by 3.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$