Solve (x-3)(x+2)=2(x+4)

$$( x - 3 ) ( x + 2 ) = 2 ( x + 4 )$$

Solve for x

$$x = \frac{\sqrt{65} + 3}{2} \approx 5.531128874$$

$$x=\frac{3-\sqrt{65}}{2}\approx -2.531128874$$

Steps Using the Quadratic Formula

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

$$x^{2}-x-6=2\left(x+4\right)$$

Use the distributive property to multiply 2 by x+4.

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

Subtract 2x from both sides.

$$x^{2}-x-6-2x=8$$

Combine -x and -2x to get -3x.

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

Subtract 8 from both sides.

$$x^{2}-3x-6-8=0$$

Subtract 8 from -6 to get -14.

$$x^{2}-3x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-14\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$x=\frac{-\left(-3\right)±\sqrt{9+56}}{2}$$

Add 9 to 56.

$$x=\frac{-\left(-3\right)±\sqrt{65}}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{65}}{2}$$

Now solve the equation x=\frac{3±\sqrt{65}}{2} when ± is plus. Add 3 to \sqrt{65}.

$$x=\frac{\sqrt{65}+3}{2}$$

Now solve the equation x=\frac{3±\sqrt{65}}{2} when ± is minus. Subtract \sqrt{65} from 3.

$$x=\frac{3-\sqrt{65}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+3}{2} x=\frac{3-\sqrt{65}}{2}$$

Steps for Completing the Square

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

$$x^{2}-x-6=2\left(x+4\right)$$

Use the distributive property to multiply 2 by x+4.

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

Subtract 2x from both sides.

$$x^{2}-x-6-2x=8$$

Combine -x and -2x to get -3x.

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

Add 6 to both sides.

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

Add 8 and 6 to get 14.

$$x^{2}-3x=14$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=14+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=14+\frac{9}{4}$$

Add 14 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{65}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{65}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{65}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{65}}{2} x-\frac{3}{2}=-\frac{\sqrt{65}}{2}$$

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

$$x=\frac{\sqrt{65}+3}{2} x=\frac{3-\sqrt{65}}{2}$$