Solve (3t-8)^2-16=0

$$( 3 t - 8 ) ^ { 2 } - 16 = 0$$

Solve for t

$$t=4$$

$$t = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-8\right)^{2}.

$$9t^{2}-48t+64-16=0$$

Subtract 16 from 64 to get 48.

$$9t^{2}-48t+48=0$$

Divide both sides by 3.

$$3t^{2}-16t+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3t^{2}+at+bt+16. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 16=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -16.

$$a=-12 b=-4$$

Rewrite 3t^{2}-16t+16 as \left(3t^{2}-12t\right)+\left(-4t+16\right).

$$\left(3t^{2}-12t\right)+\left(-4t+16\right)$$

Factor out 3t in the first and -4 in the second group.

$$3t\left(t-4\right)-4\left(t-4\right)$$

Factor out common term t-4 by using distributive property.

$$\left(t-4\right)\left(3t-4\right)$$

To find equation solutions, solve t-4=0 and 3t-4=0.

$$t=4 t=\frac{4}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-8\right)^{2}.

$$9t^{2}-48t+64-16=0$$

Subtract 16 from 64 to get 48.

$$9t^{2}-48t+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -48 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 9\times 48}}{2\times 9}$$

Square -48.

$$t=\frac{-\left(-48\right)±\sqrt{2304-4\times 9\times 48}}{2\times 9}$$

Multiply -4 times 9.

$$t=\frac{-\left(-48\right)±\sqrt{2304-36\times 48}}{2\times 9}$$

Multiply -36 times 48.

$$t=\frac{-\left(-48\right)±\sqrt{2304-1728}}{2\times 9}$$

Add 2304 to -1728.

$$t=\frac{-\left(-48\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$t=\frac{-\left(-48\right)±24}{2\times 9}$$

The opposite of -48 is 48.

$$t=\frac{48±24}{2\times 9}$$

Multiply 2 times 9.

$$t=\frac{48±24}{18}$$

Now solve the equation t=\frac{48±24}{18} when ± is plus. Add 48 to 24.

$$t=\frac{72}{18}$$

Divide 72 by 18.

$$t=4$$

Now solve the equation t=\frac{48±24}{18} when ± is minus. Subtract 24 from 48.

$$t=\frac{24}{18}$$

Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.

$$t=\frac{4}{3}$$

The equation is now solved.

$$t=4 t=\frac{4}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-8\right)^{2}.

$$9t^{2}-48t+64-16=0$$

Subtract 16 from 64 to get 48.

$$9t^{2}-48t+48=0$$

Subtract 48 from both sides. Anything subtracted from zero gives its negation.

$$9t^{2}-48t=-48$$

Divide both sides by 9.

$$\frac{9t^{2}-48t}{9}=-\frac{48}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$t^{2}+\left(-\frac{48}{9}\right)t=-\frac{48}{9}$$

Reduce the fraction \frac{-48}{9} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{16}{3}t=-\frac{48}{9}$$

Reduce the fraction \frac{-48}{9} to lowest terms by extracting and canceling out 3.

$$t^{2}-\frac{16}{3}t=-\frac{16}{3}$$

Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{16}{3}t+\left(-\frac{8}{3}\right)^{2}=-\frac{16}{3}+\left(-\frac{8}{3}\right)^{2}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{16}{3}t+\frac{64}{9}=-\frac{16}{3}+\frac{64}{9}$$

Add -\frac{16}{3} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{16}{3}t+\frac{64}{9}=\frac{16}{9}$$

Factor t^{2}-\frac{16}{3}t+\frac{64}{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}.

$$\left(t-\frac{8}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{8}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$t-\frac{8}{3}=\frac{4}{3} t-\frac{8}{3}=-\frac{4}{3}$$

Add \frac{8}{3} to both sides of the equation.

$$t=4 t=\frac{4}{3}$$