Solve p^2+5/6=p

$$\frac { p ^ { 2 } + 5 } { 6 } = p$$

Solve for p

$$p=1$$

$$p=5$$

Steps Using the Quadratic Formula

Divide each term of p^{2}+5 by 6 to get \frac{1}{6}p^{2}+\frac{5}{6}.

$$\frac{1}{6}p^{2}+\frac{5}{6}=p$$

Subtract p from both sides.

$$\frac{1}{6}p^{2}+\frac{5}{6}-p=0$$

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.

$$\frac{1}{6}p^{2}-p+\frac{5}{6}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{6} for a, -1 for b, and \frac{5}{6} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{6}\times \frac{5}{6}}}{2\times \frac{1}{6}}$$

Multiply -4 times \frac{1}{6}.

$$p=\frac{-\left(-1\right)±\sqrt{1-\frac{2}{3}\times \frac{5}{6}}}{2\times \frac{1}{6}}$$

Multiply -\frac{2}{3} times \frac{5}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$p=\frac{-\left(-1\right)±\sqrt{1-\frac{5}{9}}}{2\times \frac{1}{6}}$$

Add 1 to -\frac{5}{9}.

$$p=\frac{-\left(-1\right)±\sqrt{\frac{4}{9}}}{2\times \frac{1}{6}}$$

Take the square root of \frac{4}{9}.

$$p=\frac{-\left(-1\right)±\frac{2}{3}}{2\times \frac{1}{6}}$$

The opposite of -1 is 1.

$$p=\frac{1±\frac{2}{3}}{2\times \frac{1}{6}}$$

Multiply 2 times \frac{1}{6}.

$$p=\frac{1±\frac{2}{3}}{\frac{1}{3}}$$

Now solve the equation p=\frac{1±\frac{2}{3}}{\frac{1}{3}} when ± is plus. Add 1 to \frac{2}{3}.

$$p=\frac{\frac{5}{3}}{\frac{1}{3}}$$

Divide \frac{5}{3} by \frac{1}{3} by multiplying \frac{5}{3} by the reciprocal of \frac{1}{3}.

$$p=5$$

Now solve the equation p=\frac{1±\frac{2}{3}}{\frac{1}{3}} when ± is minus. Subtract \frac{2}{3} from 1.

$$p=\frac{\frac{1}{3}}{\frac{1}{3}}$$

Divide \frac{1}{3} by \frac{1}{3} by multiplying \frac{1}{3} by the reciprocal of \frac{1}{3}.

$$p=1$$

The equation is now solved.

$$p=5 p=1$$

Steps for Completing the Square

Divide each term of p^{2}+5 by 6 to get \frac{1}{6}p^{2}+\frac{5}{6}.

$$\frac{1}{6}p^{2}+\frac{5}{6}=p$$

Subtract p from both sides.

$$\frac{1}{6}p^{2}+\frac{5}{6}-p=0$$

Subtract \frac{5}{6} from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{6}p^{2}-p=-\frac{5}{6}$$

Multiply both sides by 6.

$$\frac{\frac{1}{6}p^{2}-p}{\frac{1}{6}}=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

Dividing by \frac{1}{6} undoes the multiplication by \frac{1}{6}.

$$p^{2}+\left(-\frac{1}{\frac{1}{6}}\right)p=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

Divide -1 by \frac{1}{6} by multiplying -1 by the reciprocal of \frac{1}{6}.

$$p^{2}-6p=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

Divide -\frac{5}{6} by \frac{1}{6} by multiplying -\frac{5}{6} by the reciprocal of \frac{1}{6}.

$$p^{2}-6p=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-6p+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$p^{2}-6p+9=-5+9$$

Add -5 to 9.

$$p^{2}-6p+9=4$$

Factor p^{2}-6p+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-3\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$p-3=2 p-3=-2$$

Add 3 to both sides of the equation.

$$p=5 p=1$$