Solve 5p^2+7=16p

$$5 p ^ { 2 } + 7 = 16 p$$

Solve for p

$$p = \frac{\sqrt{29} + 8}{5} \approx 2.677032961$$

$$p=\frac{8-\sqrt{29}}{5}\approx 0.522967039$$

Steps Using the Quadratic Formula

Subtract 16p from both sides.

$$5p^{2}+7-16p=0$$

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.

$$5p^{2}-16p+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -16 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 7}}{2\times 5}$$

Square -16.

$$p=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 7}}{2\times 5}$$

Multiply -4 times 5.

$$p=\frac{-\left(-16\right)±\sqrt{256-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$p=\frac{-\left(-16\right)±\sqrt{256-140}}{2\times 5}$$

Add 256 to -140.

$$p=\frac{-\left(-16\right)±\sqrt{116}}{2\times 5}$$

Take the square root of 116.

$$p=\frac{-\left(-16\right)±2\sqrt{29}}{2\times 5}$$

The opposite of -16 is 16.

$$p=\frac{16±2\sqrt{29}}{2\times 5}$$

Multiply 2 times 5.

$$p=\frac{16±2\sqrt{29}}{10}$$

Now solve the equation p=\frac{16±2\sqrt{29}}{10} when ± is plus. Add 16 to 2\sqrt{29}.

$$p=\frac{2\sqrt{29}+16}{10}$$

Divide 16+2\sqrt{29} by 10.

$$p=\frac{\sqrt{29}+8}{5}$$

Now solve the equation p=\frac{16±2\sqrt{29}}{10} when ± is minus. Subtract 2\sqrt{29} from 16.

$$p=\frac{16-2\sqrt{29}}{10}$$

Divide 16-2\sqrt{29} by 10.

$$p=\frac{8-\sqrt{29}}{5}$$

The equation is now solved.

$$p=\frac{\sqrt{29}+8}{5} p=\frac{8-\sqrt{29}}{5}$$

Steps for Completing the Square

Subtract 16p from both sides.

$$5p^{2}+7-16p=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$5p^{2}-16p=-7$$

Divide both sides by 5.

$$\frac{5p^{2}-16p}{5}=-\frac{7}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$p^{2}-\frac{16}{5}p=-\frac{7}{5}$$

Divide -\frac{16}{5}, the coefficient of the x term, by 2 to get -\frac{8}{5}. Then add the square of -\frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-\frac{16}{5}p+\left(-\frac{8}{5}\right)^{2}=-\frac{7}{5}+\left(-\frac{8}{5}\right)^{2}$$

Square -\frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-\frac{16}{5}p+\frac{64}{25}=-\frac{7}{5}+\frac{64}{25}$$

Add -\frac{7}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-\frac{16}{5}p+\frac{64}{25}=\frac{29}{25}$$

Factor p^{2}-\frac{16}{5}p+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-\frac{8}{5}\right)^{2}=\frac{29}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{8}{5}\right)^{2}}=\sqrt{\frac{29}{25}}$$

Simplify.

$$p-\frac{8}{5}=\frac{\sqrt{29}}{5} p-\frac{8}{5}=-\frac{\sqrt{29}}{5}$$

Add \frac{8}{5} to both sides of the equation.

$$p=\frac{\sqrt{29}+8}{5} p=\frac{8-\sqrt{29}}{5}$$