Solve (x+4)(x+8)=4x^2

$$( x + 4 ) ( x + 8 ) = 4 x ^ { 2 }$$

Solve for x

$$x=\frac{2\sqrt{33}}{3}+2\approx 5.829708431$$

$$x=-\frac{2\sqrt{33}}{3}+2\approx -1.829708431$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+4 by x+8 and combine like terms.

$$x^{2}+12x+32=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+12x+32-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+12x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 12 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\times 32}}{2\left(-3\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-3\right)\times 32}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-12±\sqrt{144+12\times 32}}{2\left(-3\right)}$$

Multiply 12 times 32.

$$x=\frac{-12±\sqrt{144+384}}{2\left(-3\right)}$$

Add 144 to 384.

$$x=\frac{-12±\sqrt{528}}{2\left(-3\right)}$$

Take the square root of 528.

$$x=\frac{-12±4\sqrt{33}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-12±4\sqrt{33}}{-6}$$

Now solve the equation x=\frac{-12±4\sqrt{33}}{-6} when ± is plus. Add -12 to 4\sqrt{33}.

$$x=\frac{4\sqrt{33}-12}{-6}$$

Divide -12+4\sqrt{33} by -6.

$$x=-\frac{2\sqrt{33}}{3}+2$$

Now solve the equation x=\frac{-12±4\sqrt{33}}{-6} when ± is minus. Subtract 4\sqrt{33} from -12.

$$x=\frac{-4\sqrt{33}-12}{-6}$$

Divide -12-4\sqrt{33} by -6.

$$x=\frac{2\sqrt{33}}{3}+2$$

The equation is now solved.

$$x=-\frac{2\sqrt{33}}{3}+2 x=\frac{2\sqrt{33}}{3}+2$$

Steps for Completing the Square

Use the distributive property to multiply x+4 by x+8 and combine like terms.

$$x^{2}+12x+32=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+12x+32-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+12x+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}+12x=-32$$

Divide both sides by -3.

$$\frac{-3x^{2}+12x}{-3}=-\frac{32}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{12}{-3}x=-\frac{32}{-3}$$

Divide 12 by -3.

$$x^{2}-4x=-\frac{32}{-3}$$

Divide -32 by -3.

$$x^{2}-4x=\frac{32}{3}$$

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

$$x^{2}-4x+\left(-2\right)^{2}=\frac{32}{3}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=\frac{32}{3}+4$$

Add \frac{32}{3} to 4.

$$x^{2}-4x+4=\frac{44}{3}$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=\frac{44}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{44}{3}}$$

Simplify.

$$x-2=\frac{2\sqrt{33}}{3} x-2=-\frac{2\sqrt{33}}{3}$$

Add 2 to both sides of the equation.

$$x=\frac{2\sqrt{33}}{3}+2 x=-\frac{2\sqrt{33}}{3}+2$$