Solve 9p^2-7=9p

$$9 p ^ { 2 } - 7 = 9 p$$

Solve for p

$$p=\frac{\sqrt{37}}{6}+\frac{1}{2}\approx 1.513793755$$

$$p=-\frac{\sqrt{37}}{6}+\frac{1}{2}\approx -0.513793755$$

Steps Using the Quadratic Formula

Subtract 9p from both sides.

$$9p^{2}-7-9p=0$$

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.

$$9p^{2}-9p-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -9 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 9\left(-7\right)}}{2\times 9}$$

Square -9.

$$p=\frac{-\left(-9\right)±\sqrt{81-4\times 9\left(-7\right)}}{2\times 9}$$

Multiply -4 times 9.

$$p=\frac{-\left(-9\right)±\sqrt{81-36\left(-7\right)}}{2\times 9}$$

Multiply -36 times -7.

$$p=\frac{-\left(-9\right)±\sqrt{81+252}}{2\times 9}$$

Add 81 to 252.

$$p=\frac{-\left(-9\right)±\sqrt{333}}{2\times 9}$$

Take the square root of 333.

$$p=\frac{-\left(-9\right)±3\sqrt{37}}{2\times 9}$$

The opposite of -9 is 9.

$$p=\frac{9±3\sqrt{37}}{2\times 9}$$

Multiply 2 times 9.

$$p=\frac{9±3\sqrt{37}}{18}$$

Now solve the equation p=\frac{9±3\sqrt{37}}{18} when ± is plus. Add 9 to 3\sqrt{37}.

$$p=\frac{3\sqrt{37}+9}{18}$$

Divide 9+3\sqrt{37} by 18.

$$p=\frac{\sqrt{37}}{6}+\frac{1}{2}$$

Now solve the equation p=\frac{9±3\sqrt{37}}{18} when ± is minus. Subtract 3\sqrt{37} from 9.

$$p=\frac{9-3\sqrt{37}}{18}$$

Divide 9-3\sqrt{37} by 18.

$$p=-\frac{\sqrt{37}}{6}+\frac{1}{2}$$

The equation is now solved.

$$p=\frac{\sqrt{37}}{6}+\frac{1}{2} p=-\frac{\sqrt{37}}{6}+\frac{1}{2}$$

Steps for Completing the Square

Subtract 9p from both sides.

$$9p^{2}-7-9p=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$9p^{2}-9p=7$$

Divide both sides by 9.

$$\frac{9p^{2}-9p}{9}=\frac{7}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$p^{2}+\left(-\frac{9}{9}\right)p=\frac{7}{9}$$

Divide -9 by 9.

$$p^{2}-p=\frac{7}{9}$$

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.

$$p^{2}-p+\left(-\frac{1}{2}\right)^{2}=\frac{7}{9}+\left(-\frac{1}{2}\right)^{2}$$

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

$$p^{2}-p+\frac{1}{4}=\frac{7}{9}+\frac{1}{4}$$

Add \frac{7}{9} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$p^{2}-p+\frac{1}{4}=\frac{37}{36}$$

Factor p^{2}-p+\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}.

$$\left(p-\frac{1}{2}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{1}{2}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$p-\frac{1}{2}=\frac{\sqrt{37}}{6} p-\frac{1}{2}=-\frac{\sqrt{37}}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$p=\frac{\sqrt{37}}{6}+\frac{1}{2} p=-\frac{\sqrt{37}}{6}+\frac{1}{2}$$