Solve (31-10t)^2-(21-10t)^2=80

$$( 31 - 10 t ) ^ { 2 } - ( 21 - 10 t ) ^ { 2 } = 80$$

Solve for t

$$t = \frac{11}{5} = 2\frac{1}{5} = 2.2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(31-10t\right)^{2}.

$$961-620t+100t^{2}-\left(21-10t\right)^{2}=80$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(21-10t\right)^{2}.

$$961-620t+100t^{2}-\left(441-420t+100t^{2}\right)=80$$

To find the opposite of 441-420t+100t^{2}, find the opposite of each term.

$$961-620t+100t^{2}-441+420t-100t^{2}=80$$

Subtract 441 from 961 to get 520.

$$520-620t+100t^{2}+420t-100t^{2}=80$$

Combine -620t and 420t to get -200t.

$$520-200t+100t^{2}-100t^{2}=80$$

Combine 100t^{2} and -100t^{2} to get 0.

$$520-200t=80$$

Subtract 520 from both sides.

$$-200t=80-520$$

Subtract 520 from 80 to get -440.

$$-200t=-440$$

Divide both sides by -200.

$$t=\frac{-440}{-200}$$

Reduce the fraction \frac{-440}{-200} to lowest terms by extracting and canceling out -40.

$$t=\frac{11}{5}$$