Solve 16=4x^2+4x

$$16 = 4 x ^ { 2 } + 4 x$$

Solve for x

$$x = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$$

$$x=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+4x=16$$

Subtract 16 from both sides.

$$4x^{2}+4x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

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

Multiply -4 times 4.

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

Multiply -16 times -16.

$$x=\frac{-4±\sqrt{16+256}}{2\times 4}$$

Add 16 to 256.

$$x=\frac{-4±\sqrt{272}}{2\times 4}$$

Take the square root of 272.

$$x=\frac{-4±4\sqrt{17}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±4\sqrt{17}}{8}$$

Now solve the equation x=\frac{-4±4\sqrt{17}}{8} when ± is plus. Add -4 to 4\sqrt{17}.

$$x=\frac{4\sqrt{17}-4}{8}$$

Divide -4+4\sqrt{17} by 8.

$$x=\frac{\sqrt{17}-1}{2}$$

Now solve the equation x=\frac{-4±4\sqrt{17}}{8} when ± is minus. Subtract 4\sqrt{17} from -4.

$$x=\frac{-4\sqrt{17}-4}{8}$$

Divide -4-4\sqrt{17} by 8.

$$x=\frac{-\sqrt{17}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+4x=16$$

Divide both sides by 4.

$$\frac{4x^{2}+4x}{4}=\frac{16}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{4}{4}x=\frac{16}{4}$$

Divide 4 by 4.

$$x^{2}+x=\frac{16}{4}$$

Divide 16 by 4.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 4 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{17}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{17}}{2} x+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{2} x=\frac{-\sqrt{17}-1}{2}$$