Solve 4[x^2+8x+16-16]-1=0

$$4 [ x ^ { 2 } + 8 x + 16 - 16 ] - 1 = 0$$

Solve for x

$$x=\frac{\sqrt{65}}{2}-4\approx 0.031128874$$

$$x=-\frac{\sqrt{65}}{2}-4\approx -8.031128874$$

Steps Using the Quadratic Formula

Subtract 16 from 16 to get 0.

$$4\left(x^{2}+8x\right)-1=0$$

Use the distributive property to multiply 4 by x^{2}+8x.

$$4x^{2}+32x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 32 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 4\left(-1\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-32±\sqrt{1024-16\left(-1\right)}}{2\times 4}$$

Multiply -16 times -1.

$$x=\frac{-32±\sqrt{1024+16}}{2\times 4}$$

Add 1024 to 16.

$$x=\frac{-32±\sqrt{1040}}{2\times 4}$$

Take the square root of 1040.

$$x=\frac{-32±4\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-32±4\sqrt{65}}{8}$$

Now solve the equation x=\frac{-32±4\sqrt{65}}{8} when ± is plus. Add -32 to 4\sqrt{65}.

$$x=\frac{4\sqrt{65}-32}{8}$$

Divide -32+4\sqrt{65} by 8.

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

Now solve the equation x=\frac{-32±4\sqrt{65}}{8} when ± is minus. Subtract 4\sqrt{65} from -32.

$$x=\frac{-4\sqrt{65}-32}{8}$$

Divide -32-4\sqrt{65} by 8.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 16 from 16 to get 0.

$$4\left(x^{2}+8x\right)-1=0$$

Use the distributive property to multiply 4 by x^{2}+8x.

$$4x^{2}+32x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$4x^{2}+32x=1$$

Divide both sides by 4.

$$\frac{4x^{2}+32x}{4}=\frac{1}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{32}{4}x=\frac{1}{4}$$

Divide 32 by 4.

$$x^{2}+8x=\frac{1}{4}$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=\frac{1}{4}+4^{2}$$

Square 4.

$$x^{2}+8x+16=\frac{1}{4}+16$$

Add \frac{1}{4} to 16.

$$x^{2}+8x+16=\frac{65}{4}$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+4=\frac{\sqrt{65}}{2} x+4=-\frac{\sqrt{65}}{2}$$

Subtract 4 from both sides of the equation.

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