Solve (40-4t)^2+(30-3t)^2=30

$$( 40 - 4 t ) ^ { 2 } + ( 30 - 3 t ) ^ { 2 } = 30$$

Solve for t

$$t=\frac{\sqrt{30}}{5}+10\approx 11.095445115$$

$$t=-\frac{\sqrt{30}}{5}+10\approx 8.904554885$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-4t\right)^{2}.

$$1600-320t+16t^{2}+\left(30-3t\right)^{2}=30$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(30-3t\right)^{2}.

$$1600-320t+16t^{2}+900-180t+9t^{2}=30$$

Add 1600 and 900 to get 2500.

$$2500-320t+16t^{2}-180t+9t^{2}=30$$

Combine -320t and -180t to get -500t.

$$2500-500t+16t^{2}+9t^{2}=30$$

Combine 16t^{2} and 9t^{2} to get 25t^{2}.

$$2500-500t+25t^{2}=30$$

Subtract 30 from both sides.

$$2500-500t+25t^{2}-30=0$$

Subtract 30 from 2500 to get 2470.

$$2470-500t+25t^{2}=0$$

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.

$$25t^{2}-500t+2470=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -500 for b, and 2470 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-500\right)±\sqrt{\left(-500\right)^{2}-4\times 25\times 2470}}{2\times 25}$$

Square -500.

$$t=\frac{-\left(-500\right)±\sqrt{250000-4\times 25\times 2470}}{2\times 25}$$

Multiply -4 times 25.

$$t=\frac{-\left(-500\right)±\sqrt{250000-100\times 2470}}{2\times 25}$$

Multiply -100 times 2470.

$$t=\frac{-\left(-500\right)±\sqrt{250000-247000}}{2\times 25}$$

Add 250000 to -247000.

$$t=\frac{-\left(-500\right)±\sqrt{3000}}{2\times 25}$$

Take the square root of 3000.

$$t=\frac{-\left(-500\right)±10\sqrt{30}}{2\times 25}$$

The opposite of -500 is 500.

$$t=\frac{500±10\sqrt{30}}{2\times 25}$$

Multiply 2 times 25.

$$t=\frac{500±10\sqrt{30}}{50}$$

Now solve the equation t=\frac{500±10\sqrt{30}}{50} when ± is plus. Add 500 to 10\sqrt{30}.

$$t=\frac{10\sqrt{30}+500}{50}$$

Divide 500+10\sqrt{30} by 50.

$$t=\frac{\sqrt{30}}{5}+10$$

Now solve the equation t=\frac{500±10\sqrt{30}}{50} when ± is minus. Subtract 10\sqrt{30} from 500.

$$t=\frac{500-10\sqrt{30}}{50}$$

Divide 500-10\sqrt{30} by 50.

$$t=-\frac{\sqrt{30}}{5}+10$$

The equation is now solved.

$$t=\frac{\sqrt{30}}{5}+10 t=-\frac{\sqrt{30}}{5}+10$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-4t\right)^{2}.

$$1600-320t+16t^{2}+\left(30-3t\right)^{2}=30$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(30-3t\right)^{2}.

$$1600-320t+16t^{2}+900-180t+9t^{2}=30$$

Add 1600 and 900 to get 2500.

$$2500-320t+16t^{2}-180t+9t^{2}=30$$

Combine -320t and -180t to get -500t.

$$2500-500t+16t^{2}+9t^{2}=30$$

Combine 16t^{2} and 9t^{2} to get 25t^{2}.

$$2500-500t+25t^{2}=30$$

Subtract 2500 from both sides.

$$-500t+25t^{2}=30-2500$$

Subtract 2500 from 30 to get -2470.

$$-500t+25t^{2}=-2470$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$25t^{2}-500t=-2470$$

Divide both sides by 25.

$$\frac{25t^{2}-500t}{25}=-\frac{2470}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$t^{2}+\left(-\frac{500}{25}\right)t=-\frac{2470}{25}$$

Divide -500 by 25.

$$t^{2}-20t=-\frac{2470}{25}$$

Reduce the fraction \frac{-2470}{25} to lowest terms by extracting and canceling out 5.

$$t^{2}-20t=-\frac{494}{5}$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-20t+\left(-10\right)^{2}=-\frac{494}{5}+\left(-10\right)^{2}$$

Square -10.

$$t^{2}-20t+100=-\frac{494}{5}+100$$

Add -\frac{494}{5} to 100.

$$t^{2}-20t+100=\frac{6}{5}$$

Factor t^{2}-20t+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-10\right)^{2}=\frac{6}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-10\right)^{2}}=\sqrt{\frac{6}{5}}$$

Simplify.

$$t-10=\frac{\sqrt{30}}{5} t-10=-\frac{\sqrt{30}}{5}$$

Add 10 to both sides of the equation.

$$t=\frac{\sqrt{30}}{5}+10 t=-\frac{\sqrt{30}}{5}+10$$