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

$$( 2 x + 1 ) ( x + 3 ) - 9 = ( 3 x - 2 ) ( x + 2 )$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

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

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

Subtract 9 from 3 to get -6.

$$2x^{2}+7x-6=\left(3x-2\right)\left(x+2\right)$$

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

$$2x^{2}+7x-6=3x^{2}+4x-4$$

Subtract 3x^{2} from both sides.

$$2x^{2}+7x-6-3x^{2}=4x-4$$

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

$$-x^{2}+7x-6=4x-4$$

Subtract 4x from both sides.

$$-x^{2}+7x-6-4x=-4$$

Combine 7x and -4x to get 3x.

$$-x^{2}+3x-6=-4$$

Add 4 to both sides.

$$-x^{2}+3x-6+4=0$$

Add -6 and 4 to get -2.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

$$x=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±1}{-2}$$

Now solve the equation x=\frac{-3±1}{-2} when ± is plus. Add -3 to 1.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-3±1}{-2} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

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

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

Subtract 9 from 3 to get -6.

$$2x^{2}+7x-6=\left(3x-2\right)\left(x+2\right)$$

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

$$2x^{2}+7x-6=3x^{2}+4x-4$$

Subtract 3x^{2} from both sides.

$$2x^{2}+7x-6-3x^{2}=4x-4$$

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

$$-x^{2}+7x-6=4x-4$$

Subtract 4x from both sides.

$$-x^{2}+7x-6-4x=-4$$

Combine 7x and -4x to get 3x.

$$-x^{2}+3x-6=-4$$

Add 6 to both sides.

$$-x^{2}+3x=-4+6$$

Add -4 and 6 to get 2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

$$x^{2}-3x=\frac{2}{-1}$$

Divide 2 by -1.

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

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.

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

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

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

Add -2 to \frac{9}{4}.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

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

$$x=2 x=1$$