Solve 71=1/3(-2-h)^2-4

$$71 = \frac { 1 } { 3 } ( - 2 - h ) ^ { 2 } - 4$$

Solve for h

$$h=13$$

$$h=-17$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2-h\right)^{2}.

$$71=\frac{1}{3}\left(4+4h+h^{2}\right)-4$$

Use the distributive property to multiply \frac{1}{3} by 4+4h+h^{2}.

$$71=\frac{4}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}-4$$

Subtract 4 from \frac{4}{3} to get -\frac{8}{3}.

$$71=-\frac{8}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{8}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}=71$$

Subtract 71 from both sides.

$$-\frac{8}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}-71=0$$

Subtract 71 from -\frac{8}{3} to get -\frac{221}{3}.

$$-\frac{221}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}=0$$

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.

$$\frac{1}{3}h^{2}+\frac{4}{3}h-\frac{221}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, \frac{4}{3} for b, and -\frac{221}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{-\frac{4}{3}±\sqrt{\left(\frac{4}{3}\right)^{2}-4\times \frac{1}{3}\left(-\frac{221}{3}\right)}}{2\times \frac{1}{3}}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$h=\frac{-\frac{4}{3}±\sqrt{\frac{16}{9}-4\times \frac{1}{3}\left(-\frac{221}{3}\right)}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$h=\frac{-\frac{4}{3}±\sqrt{\frac{16}{9}-\frac{4}{3}\left(-\frac{221}{3}\right)}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times -\frac{221}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$h=\frac{-\frac{4}{3}±\sqrt{\frac{16+884}{9}}}{2\times \frac{1}{3}}$$

Add \frac{16}{9} to \frac{884}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$h=\frac{-\frac{4}{3}±\sqrt{100}}{2\times \frac{1}{3}}$$

Take the square root of 100.

$$h=\frac{-\frac{4}{3}±10}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$h=\frac{-\frac{4}{3}±10}{\frac{2}{3}}$$

Now solve the equation h=\frac{-\frac{4}{3}±10}{\frac{2}{3}} when ± is plus. Add -\frac{4}{3} to 10.

$$h=\frac{\frac{26}{3}}{\frac{2}{3}}$$

Divide \frac{26}{3} by \frac{2}{3} by multiplying \frac{26}{3} by the reciprocal of \frac{2}{3}.

$$h=13$$

Now solve the equation h=\frac{-\frac{4}{3}±10}{\frac{2}{3}} when ± is minus. Subtract 10 from -\frac{4}{3}.

$$h=-\frac{\frac{34}{3}}{\frac{2}{3}}$$

Divide -\frac{34}{3} by \frac{2}{3} by multiplying -\frac{34}{3} by the reciprocal of \frac{2}{3}.

$$h=-17$$

The equation is now solved.

$$h=13 h=-17$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-2-h\right)^{2}.

$$71=\frac{1}{3}\left(4+4h+h^{2}\right)-4$$

Use the distributive property to multiply \frac{1}{3} by 4+4h+h^{2}.

$$71=\frac{4}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}-4$$

Subtract 4 from \frac{4}{3} to get -\frac{8}{3}.

$$71=-\frac{8}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$-\frac{8}{3}+\frac{4}{3}h+\frac{1}{3}h^{2}=71$$

Add \frac{8}{3} to both sides.

$$\frac{4}{3}h+\frac{1}{3}h^{2}=71+\frac{8}{3}$$

Add 71 and \frac{8}{3} to get \frac{221}{3}.

$$\frac{4}{3}h+\frac{1}{3}h^{2}=\frac{221}{3}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{3}h^{2}+\frac{4}{3}h=\frac{221}{3}$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}h^{2}+\frac{4}{3}h}{\frac{1}{3}}=\frac{\frac{221}{3}}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$h^{2}+\frac{\frac{4}{3}}{\frac{1}{3}}h=\frac{\frac{221}{3}}{\frac{1}{3}}$$

Divide \frac{4}{3} by \frac{1}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{1}{3}.

$$h^{2}+4h=\frac{\frac{221}{3}}{\frac{1}{3}}$$

Divide \frac{221}{3} by \frac{1}{3} by multiplying \frac{221}{3} by the reciprocal of \frac{1}{3}.

$$h^{2}+4h=221$$

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

$$h^{2}+4h+2^{2}=221+2^{2}$$

Square 2.

$$h^{2}+4h+4=221+4$$

Add 221 to 4.

$$h^{2}+4h+4=225$$

Factor h^{2}+4h+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}.

$$\left(h+2\right)^{2}=225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(h+2\right)^{2}}=\sqrt{225}$$

Simplify.

$$h+2=15 h+2=-15$$

Subtract 2 from both sides of the equation.

$$h=13 h=-17$$