Solve 2x(6+2x)=56

$$2 x ( 6 + 2 x ) = 56$$

Solve for x

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

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by 6+2x.

$$12x+4x^{2}=56$$

Subtract 56 from both sides.

$$12x+4x^{2}-56=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4x^{2}+12x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 12 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\left(-56\right)}}{2\times 4}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 4\left(-56\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-12±\sqrt{144-16\left(-56\right)}}{2\times 4}$$

Multiply -16 times -56.

$$x=\frac{-12±\sqrt{144+896}}{2\times 4}$$

Add 144 to 896.

$$x=\frac{-12±\sqrt{1040}}{2\times 4}$$

Take the square root of 1040.

$$x=\frac{-12±4\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±4\sqrt{65}}{8}$$

Now solve the equation x=\frac{-12±4\sqrt{65}}{8} when ± is plus. Add -12 to 4\sqrt{65}.

$$x=\frac{4\sqrt{65}-12}{8}$$

Divide -12+4\sqrt{65} by 8.

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

Now solve the equation x=\frac{-12±4\sqrt{65}}{8} when ± is minus. Subtract 4\sqrt{65} from -12.

$$x=\frac{-4\sqrt{65}-12}{8}$$

Divide -12-4\sqrt{65} by 8.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 2x by 6+2x.

$$12x+4x^{2}=56$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4x^{2}+12x=56$$

Divide both sides by 4.

$$\frac{4x^{2}+12x}{4}=\frac{56}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{12}{4}x=\frac{56}{4}$$

Divide 12 by 4.

$$x^{2}+3x=\frac{56}{4}$$

Divide 56 by 4.

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

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.

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

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

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