Solve (3t-5)^2=225

$$( 3 t - 5 ) ^ { 2 } = 225$$

Solve for t

$$t = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

$$t = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-5\right)^{2}.

$$9t^{2}-30t+25=225$$

Subtract 225 from both sides.

$$9t^{2}-30t+25-225=0$$

Subtract 225 from 25 to get -200.

$$9t^{2}-30t-200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9t^{2}+at+bt-200. To find a and b, set up a system to be solved.

$$a+b=-30 ab=9\left(-200\right)=-1800$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1800.

$$1,-1800 2,-900 3,-600 4,-450 5,-360 6,-300 8,-225 9,-200 10,-180 12,-150 15,-120 18,-100 20,-90 24,-75 25,-72 30,-60 36,-50 40,-45$$

Calculate the sum for each pair.

$$1-1800=-1799 2-900=-898 3-600=-597 4-450=-446 5-360=-355 6-300=-294 8-225=-217 9-200=-191 10-180=-170 12-150=-138 15-120=-105 18-100=-82 20-90=-70 24-75=-51 25-72=-47 30-60=-30 36-50=-14 40-45=-5$$

The solution is the pair that gives sum -30.

$$a=-60 b=30$$

Rewrite 9t^{2}-30t-200 as \left(9t^{2}-60t\right)+\left(30t-200\right).

$$\left(9t^{2}-60t\right)+\left(30t-200\right)$$

Factor out 3t in the first and 10 in the second group.

$$3t\left(3t-20\right)+10\left(3t-20\right)$$

Factor out common term 3t-20 by using distributive property.

$$\left(3t-20\right)\left(3t+10\right)$$

To find equation solutions, solve 3t-20=0 and 3t+10=0.

$$t=\frac{20}{3} t=-\frac{10}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-5\right)^{2}.

$$9t^{2}-30t+25=225$$

Subtract 225 from both sides.

$$9t^{2}-30t+25-225=0$$

Subtract 225 from 25 to get -200.

$$9t^{2}-30t-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\left(-200\right)}}{2\times 9}$$

Square -30.

$$t=\frac{-\left(-30\right)±\sqrt{900-4\times 9\left(-200\right)}}{2\times 9}$$

Multiply -4 times 9.

$$t=\frac{-\left(-30\right)±\sqrt{900-36\left(-200\right)}}{2\times 9}$$

Multiply -36 times -200.

$$t=\frac{-\left(-30\right)±\sqrt{900+7200}}{2\times 9}$$

Add 900 to 7200.

$$t=\frac{-\left(-30\right)±\sqrt{8100}}{2\times 9}$$

Take the square root of 8100.

$$t=\frac{-\left(-30\right)±90}{2\times 9}$$

The opposite of -30 is 30.

$$t=\frac{30±90}{2\times 9}$$

Multiply 2 times 9.

$$t=\frac{30±90}{18}$$

Now solve the equation t=\frac{30±90}{18} when ± is plus. Add 30 to 90.

$$t=\frac{120}{18}$$

Reduce the fraction \frac{120}{18} to lowest terms by extracting and canceling out 6.

$$t=\frac{20}{3}$$

Now solve the equation t=\frac{30±90}{18} when ± is minus. Subtract 90 from 30.

$$t=-\frac{60}{18}$$

Reduce the fraction \frac{-60}{18} to lowest terms by extracting and canceling out 6.

$$t=-\frac{10}{3}$$

The equation is now solved.

$$t=\frac{20}{3} t=-\frac{10}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-5\right)^{2}.

$$9t^{2}-30t+25=225$$

Subtract 25 from both sides.

$$9t^{2}-30t=225-25$$

Subtract 25 from 225 to get 200.

$$9t^{2}-30t=200$$

Divide both sides by 9.

$$\frac{9t^{2}-30t}{9}=\frac{200}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$t^{2}+\left(-\frac{30}{9}\right)t=\frac{200}{9}$$

Reduce the fraction \frac{-30}{9} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{10}{3}t=\frac{200}{9}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{10}{3}t+\left(-\frac{5}{3}\right)^{2}=\frac{200}{9}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{10}{3}t+\frac{25}{9}=\frac{200+25}{9}$$

Add \frac{200}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{10}{3}t+\frac{25}{9}=25$$

Factor t^{2}-\frac{10}{3}t+\frac{25}{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}.

$$\left(t-\frac{5}{3}\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{3}\right)^{2}}=\sqrt{25}$$

Simplify.

$$t-\frac{5}{3}=5 t-\frac{5}{3}=-5$$

Add \frac{5}{3} to both sides of the equation.

$$t=\frac{20}{3} t=-\frac{10}{3}$$