Solve (3x-3)left(x+6right)=2xleft(x+3right)

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

Solve for x

$$x = \frac{3 \sqrt{17} - 9}{2} \approx 1.684658438$$

$$x=\frac{-3\sqrt{17}-9}{2}\approx -10.684658438$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-3 by x+6 and combine like terms.

$$3x^{2}+15x-18=2x\left(x+3\right)$$

Use the distributive property to multiply 2x by x+3.

$$3x^{2}+15x-18=2x^{2}+6x$$

Subtract 2x^{2} from both sides.

$$3x^{2}+15x-18-2x^{2}=6x$$

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

$$x^{2}+15x-18=6x$$

Subtract 6x from both sides.

$$x^{2}+15x-18-6x=0$$

Combine 15x and -6x to get 9x.

$$x^{2}+9x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-18\right)}}{2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-9±\sqrt{81+72}}{2}$$

Add 81 to 72.

$$x=\frac{-9±\sqrt{153}}{2}$$

Take the square root of 153.

$$x=\frac{-9±3\sqrt{17}}{2}$$

Now solve the equation x=\frac{-9±3\sqrt{17}}{2} when ± is plus. Add -9 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}-9}{2}$$

Now solve the equation x=\frac{-9±3\sqrt{17}}{2} when ± is minus. Subtract 3\sqrt{17} from -9.

$$x=\frac{-3\sqrt{17}-9}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{17}-9}{2} x=\frac{-3\sqrt{17}-9}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-3 by x+6 and combine like terms.

$$3x^{2}+15x-18=2x\left(x+3\right)$$

Use the distributive property to multiply 2x by x+3.

$$3x^{2}+15x-18=2x^{2}+6x$$

Subtract 2x^{2} from both sides.

$$3x^{2}+15x-18-2x^{2}=6x$$

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

$$x^{2}+15x-18=6x$$

Subtract 6x from both sides.

$$x^{2}+15x-18-6x=0$$

Combine 15x and -6x to get 9x.

$$x^{2}+9x-18=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$x^{2}+9x=18$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=18+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=18+\frac{81}{4}$$

Add 18 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{153}{4}$$

Factor x^{2}+9x+\frac{81}{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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{153}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{3\sqrt{17}}{2} x+\frac{9}{2}=-\frac{3\sqrt{17}}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=\frac{3\sqrt{17}-9}{2} x=\frac{-3\sqrt{17}-9}{2}$$