Solve left(16pright)^2+8p-8=0

$${ \left(16p \right) }^{ 2 } +8p-8 = 0$$

Solve for p

$$p=\frac{\sqrt{129}-1}{64}\approx 0.161840886$$

$$p=\frac{-\sqrt{129}-1}{64}\approx -0.193090886$$

Steps Using the Quadratic Formula

Expand \left(16p\right)^{2}.

$$16^{2}p^{2}+8p-8=0$$

Calculate 16 to the power of 2 and get 256.

$$256p^{2}+8p-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 256 for a, 8 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-8±\sqrt{8^{2}-4\times 256\left(-8\right)}}{2\times 256}$$

Square 8.

$$p=\frac{-8±\sqrt{64-4\times 256\left(-8\right)}}{2\times 256}$$

Multiply -4 times 256.

$$p=\frac{-8±\sqrt{64-1024\left(-8\right)}}{2\times 256}$$

Multiply -1024 times -8.

$$p=\frac{-8±\sqrt{64+8192}}{2\times 256}$$

Add 64 to 8192.

$$p=\frac{-8±\sqrt{8256}}{2\times 256}$$

Take the square root of 8256.

$$p=\frac{-8±8\sqrt{129}}{2\times 256}$$

Multiply 2 times 256.

$$p=\frac{-8±8\sqrt{129}}{512}$$

Now solve the equation p=\frac{-8±8\sqrt{129}}{512} when ± is plus. Add -8 to 8\sqrt{129}.

$$p=\frac{8\sqrt{129}-8}{512}$$

Divide -8+8\sqrt{129} by 512.

$$p=\frac{\sqrt{129}-1}{64}$$

Now solve the equation p=\frac{-8±8\sqrt{129}}{512} when ± is minus. Subtract 8\sqrt{129} from -8.

$$p=\frac{-8\sqrt{129}-8}{512}$$

Divide -8-8\sqrt{129} by 512.

$$p=\frac{-\sqrt{129}-1}{64}$$

The equation is now solved.

$$p=\frac{\sqrt{129}-1}{64} p=\frac{-\sqrt{129}-1}{64}$$

Steps for Completing the Square

Expand \left(16p\right)^{2}.

$$16^{2}p^{2}+8p-8=0$$

Calculate 16 to the power of 2 and get 256.

$$256p^{2}+8p-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$256p^{2}+8p=8$$

Divide both sides by 256.

$$\frac{256p^{2}+8p}{256}=\frac{8}{256}$$

Dividing by 256 undoes the multiplication by 256.

$$p^{2}+\frac{8}{256}p=\frac{8}{256}$$

Reduce the fraction \frac{8}{256} to lowest terms by extracting and canceling out 8.

$$p^{2}+\frac{1}{32}p=\frac{8}{256}$$

Reduce the fraction \frac{8}{256} to lowest terms by extracting and canceling out 8.

$$p^{2}+\frac{1}{32}p=\frac{1}{32}$$

Divide \frac{1}{32}, the coefficient of the x term, by 2 to get \frac{1}{64}. Then add the square of \frac{1}{64} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}+\frac{1}{32}p+\left(\frac{1}{64}\right)^{2}=\frac{1}{32}+\left(\frac{1}{64}\right)^{2}$$

Square \frac{1}{64} by squaring both the numerator and the denominator of the fraction.

$$p^{2}+\frac{1}{32}p+\frac{1}{4096}=\frac{1}{32}+\frac{1}{4096}$$

Add \frac{1}{32} to \frac{1}{4096} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}+\frac{1}{32}p+\frac{1}{4096}=\frac{129}{4096}$$

Factor p^{2}+\frac{1}{32}p+\frac{1}{4096}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p+\frac{1}{64}\right)^{2}=\frac{129}{4096}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p+\frac{1}{64}\right)^{2}}=\sqrt{\frac{129}{4096}}$$

Simplify.

$$p+\frac{1}{64}=\frac{\sqrt{129}}{64} p+\frac{1}{64}=-\frac{\sqrt{129}}{64}$$

Subtract \frac{1}{64} from both sides of the equation.

$$p=\frac{\sqrt{129}-1}{64} p=\frac{-\sqrt{129}-1}{64}$$