Solve 72=8/(y-3)^2

$$72 = \frac { 8 } { ( y - 3 ) ^ { 2 } }$$

Solve for y

$$y = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$y = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(y-3\right)^{2}.

$$72\left(y-3\right)^{2}=8$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$72\left(y^{2}-6y+9\right)=8$$

Use the distributive property to multiply 72 by y^{2}-6y+9.

$$72y^{2}-432y+648=8$$

Subtract 8 from both sides.

$$72y^{2}-432y+648-8=0$$

Subtract 8 from 648 to get 640.

$$72y^{2}-432y+640=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 72 for a, -432 for b, and 640 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-432\right)±\sqrt{\left(-432\right)^{2}-4\times 72\times 640}}{2\times 72}$$

Square -432.

$$y=\frac{-\left(-432\right)±\sqrt{186624-4\times 72\times 640}}{2\times 72}$$

Multiply -4 times 72.

$$y=\frac{-\left(-432\right)±\sqrt{186624-288\times 640}}{2\times 72}$$

Multiply -288 times 640.

$$y=\frac{-\left(-432\right)±\sqrt{186624-184320}}{2\times 72}$$

Add 186624 to -184320.

$$y=\frac{-\left(-432\right)±\sqrt{2304}}{2\times 72}$$

Take the square root of 2304.

$$y=\frac{-\left(-432\right)±48}{2\times 72}$$

The opposite of -432 is 432.

$$y=\frac{432±48}{2\times 72}$$

Multiply 2 times 72.

$$y=\frac{432±48}{144}$$

Now solve the equation y=\frac{432±48}{144} when ± is plus. Add 432 to 48.

$$y=\frac{480}{144}$$

Reduce the fraction \frac{480}{144} to lowest terms by extracting and canceling out 48.

$$y=\frac{10}{3}$$

Now solve the equation y=\frac{432±48}{144} when ± is minus. Subtract 48 from 432.

$$y=\frac{384}{144}$$

Reduce the fraction \frac{384}{144} to lowest terms by extracting and canceling out 48.

$$y=\frac{8}{3}$$

The equation is now solved.

$$y=\frac{10}{3} y=\frac{8}{3}$$

Steps for Completing the Square

Variable y cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(y-3\right)^{2}.

$$72\left(y-3\right)^{2}=8$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$72\left(y^{2}-6y+9\right)=8$$

Use the distributive property to multiply 72 by y^{2}-6y+9.

$$72y^{2}-432y+648=8$$

Subtract 648 from both sides.

$$72y^{2}-432y=8-648$$

Subtract 648 from 8 to get -640.

$$72y^{2}-432y=-640$$

Divide both sides by 72.

$$\frac{72y^{2}-432y}{72}=-\frac{640}{72}$$

Dividing by 72 undoes the multiplication by 72.

$$y^{2}+\left(-\frac{432}{72}\right)y=-\frac{640}{72}$$

Divide -432 by 72.

$$y^{2}-6y=-\frac{640}{72}$$

Reduce the fraction \frac{-640}{72} to lowest terms by extracting and canceling out 8.

$$y^{2}-6y=-\frac{80}{9}$$

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.

$$y^{2}-6y+\left(-3\right)^{2}=-\frac{80}{9}+\left(-3\right)^{2}$$

Square -3.

$$y^{2}-6y+9=-\frac{80}{9}+9$$

Add -\frac{80}{9} to 9.

$$y^{2}-6y+9=\frac{1}{9}$$

Factor y^{2}-6y+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}.

$$\left(y-3\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-3\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$y-3=\frac{1}{3} y-3=-\frac{1}{3}$$

Add 3 to both sides of the equation.

$$y=\frac{10}{3} y=\frac{8}{3}$$