Solve 3x(x+2)-9=2x-8

$$3 x ( x + 2 ) - 9 = 2 x - 8$$

Solve for x

$$x=\frac{\sqrt{7}-2}{3}\approx 0.215250437$$

$$x=\frac{-\sqrt{7}-2}{3}\approx -1.54858377$$

Steps Using the Quadratic Formula

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

$$3x^{2}+6x-9=2x-8$$

Subtract 2x from both sides.

$$3x^{2}+6x-9-2x=-8$$

Combine 6x and -2x to get 4x.

$$3x^{2}+4x-9=-8$$

Add 8 to both sides.

$$3x^{2}+4x-9+8=0$$

Add -9 and 8 to get -1.

$$3x^{2}+4x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-1\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12\left(-1\right)}}{2\times 3}$$

Multiply -12 times -1.

$$x=\frac{-4±\sqrt{16+12}}{2\times 3}$$

Add 16 to 12.

$$x=\frac{-4±\sqrt{28}}{2\times 3}$$

Take the square root of 28.

$$x=\frac{-4±2\sqrt{7}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-4±2\sqrt{7}}{6}$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{6} when ± is plus. Add -4 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-4}{6}$$

Divide -4+2\sqrt{7} by 6.

$$x=\frac{\sqrt{7}-2}{3}$$

Now solve the equation x=\frac{-4±2\sqrt{7}}{6} when ± is minus. Subtract 2\sqrt{7} from -4.

$$x=\frac{-2\sqrt{7}-4}{6}$$

Divide -4-2\sqrt{7} by 6.

$$x=\frac{-\sqrt{7}-2}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$

Steps for Completing the Square

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

$$3x^{2}+6x-9=2x-8$$

Subtract 2x from both sides.

$$3x^{2}+6x-9-2x=-8$$

Combine 6x and -2x to get 4x.

$$3x^{2}+4x-9=-8$$

Add 9 to both sides.

$$3x^{2}+4x=-8+9$$

Add -8 and 9 to get 1.

$$3x^{2}+4x=1$$

Divide both sides by 3.

$$\frac{3x^{2}+4x}{3}=\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{4}{3}x=\frac{1}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{1}{3}+\left(\frac{2}{3}\right)^{2}$$

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

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{3}+\frac{4}{9}$$

Add \frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{7}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{7}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{\sqrt{7}}{3} x+\frac{2}{3}=-\frac{\sqrt{7}}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=\frac{\sqrt{7}-2}{3} x=\frac{-\sqrt{7}-2}{3}$$