Solve (20t)^2+(90-30t)^2=50^2

$$( 20 t ) ^ { 2 } + ( 90 - 30 t ) ^ { 2 } = 50 ^ { 2 }$$

Solve for t

$$t=2$$

$$t = \frac{28}{13} = 2\frac{2}{13} \approx 2.153846154$$

Steps Using the Quadratic Formula

Expand \left(20t\right)^{2}.

$$20^{2}t^{2}+\left(90-30t\right)^{2}=50^{2}$$

Calculate 20 to the power of 2 and get 400.

$$400t^{2}+\left(90-30t\right)^{2}=50^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(90-30t\right)^{2}.

$$400t^{2}[phone]t+900t^{2}=50^{2}$$

Combine 400t^{2} and 900t^{2} to get 1300t^{2}.

$$1300t^{2}[phone]t=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$1300t^{2}[phone]t=2500$$

Subtract 2500 from both sides.

$$1300t^{2}[phone]t-2500=0$$

Subtract 2500 from 8100 to get 5600.

$$1300t^{2}[phone]t=0$$

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.

$$1300t^{2}-5400t+5600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1300 for a, -5400 for b, and 5600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-5400\right)±\sqrt{\left(-5400\right)^{2}-4\times 1300\times 5600}}{2\times 1300}$$

Square -5400.

$$t=\frac{-\left(-5400\right)±\sqrt{29160000-4\times 1300\times 5600}}{2\times 1300}$$

Multiply -4 times 1300.

$$t=\frac{-\left(-5400\right)±\sqrt{29160000-5200\times 5600}}{2\times 1300}$$

Multiply -5200 times 5600.

$$t=\frac{-\left(-5400\right)±\sqrt{29160000-29120000}}{2\times 1300}$$

Add 29160000 to -29120000.

$$t=\frac{-\left(-5400\right)±\sqrt{40000}}{2\times 1300}$$

Take the square root of 40000.

$$t=\frac{-\left(-5400\right)±200}{2\times 1300}$$

The opposite of -5400 is 5400.

$$t=\frac{5400±200}{2\times 1300}$$

Multiply 2 times 1300.

$$t=\frac{5400±200}{2600}$$

Now solve the equation t=\frac{5400±200}{2600} when ± is plus. Add 5400 to 200.

$$t=\frac{5600}{2600}$$

Reduce the fraction \frac{5600}{2600} to lowest terms by extracting and canceling out 200.

$$t=\frac{28}{13}$$

Now solve the equation t=\frac{5400±200}{2600} when ± is minus. Subtract 200 from 5400.

$$t=\frac{5200}{2600}$$

Divide 5200 by 2600.

$$t=2$$

The equation is now solved.

$$t=\frac{28}{13} t=2$$

Steps for Completing the Square

Expand \left(20t\right)^{2}.

$$20^{2}t^{2}+\left(90-30t\right)^{2}=50^{2}$$

Calculate 20 to the power of 2 and get 400.

$$400t^{2}+\left(90-30t\right)^{2}=50^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(90-30t\right)^{2}.

$$400t^{2}[phone]t+900t^{2}=50^{2}$$

Combine 400t^{2} and 900t^{2} to get 1300t^{2}.

$$1300t^{2}[phone]t=50^{2}$$

Calculate 50 to the power of 2 and get 2500.

$$1300t^{2}[phone]t=2500$$

Subtract 8100 from both sides.

$$1300t^{2}-5400t=2500-8100$$

Subtract 8100 from 2500 to get -5600.

$$1300t^{2}-5400t=-5600$$

Divide both sides by 1300.

$$\frac{1300t^{2}-5400t}{1300}=-\frac{5600}{1300}$$

Dividing by 1300 undoes the multiplication by 1300.

$$t^{2}+\left(-\frac{5400}{1300}\right)t=-\frac{5600}{1300}$$

Reduce the fraction \frac{-5400}{1300} to lowest terms by extracting and canceling out 100.

$$t^{2}-\frac{54}{13}t=-\frac{5600}{1300}$$

Reduce the fraction \frac{-5600}{1300} to lowest terms by extracting and canceling out 100.

$$t^{2}-\frac{54}{13}t=-\frac{56}{13}$$

Divide -\frac{54}{13}, the coefficient of the x term, by 2 to get -\frac{27}{13}. Then add the square of -\frac{27}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{54}{13}t+\left(-\frac{27}{13}\right)^{2}=-\frac{56}{13}+\left(-\frac{27}{13}\right)^{2}$$

Square -\frac{27}{13} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{54}{13}t+\frac{729}{169}=-\frac{56}{13}+\frac{729}{169}$$

Add -\frac{56}{13} to \frac{729}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{54}{13}t+\frac{729}{169}=\frac{1}{169}$$

Factor t^{2}-\frac{54}{13}t+\frac{729}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{27}{13}\right)^{2}=\frac{1}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{27}{13}\right)^{2}}=\sqrt{\frac{1}{169}}$$

Simplify.

$$t-\frac{27}{13}=\frac{1}{13} t-\frac{27}{13}=-\frac{1}{13}$$

Add \frac{27}{13} to both sides of the equation.

$$t=\frac{28}{13} t=2$$