Solve (3x-2)left(x+1right)=9

$$(3x-2) \left( x+1 \right) = 9$$

Solve for x

$$x = \frac{\sqrt{133} - 1}{6} \approx 1.755427099$$

$$x=\frac{-\sqrt{133}-1}{6}\approx -2.088760432$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-2 by x+1 and combine like terms.

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

Subtract 9 from both sides.

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

Subtract 9 from -2 to get -11.

$$3x^{2}+x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-11\right)}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-11\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-11\right)}}{2\times 3}$$

Multiply -12 times -11.

$$x=\frac{-1±\sqrt{1+132}}{2\times 3}$$

Add 1 to 132.

$$x=\frac{-1±\sqrt{133}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±\sqrt{133}}{6}$$

Now solve the equation x=\frac{-1±\sqrt{133}}{6} when ± is plus. Add -1 to \sqrt{133}.

$$x=\frac{\sqrt{133}-1}{6}$$

Now solve the equation x=\frac{-1±\sqrt{133}}{6} when ± is minus. Subtract \sqrt{133} from -1.

$$x=\frac{-\sqrt{133}-1}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-2 by x+1 and combine like terms.

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

Add 2 to both sides.

$$3x^{2}+x=9+2$$

Add 9 and 2 to get 11.

$$3x^{2}+x=11$$

Divide both sides by 3.

$$\frac{3x^{2}+x}{3}=\frac{11}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{1}{3}x=\frac{11}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{11}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{11}{3}+\frac{1}{36}$$

Add \frac{11}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{133}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{133}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{133}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{133}}{6} x+\frac{1}{6}=-\frac{\sqrt{133}}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$