Solve (16-2t)^2=5t^2+36

$$( 16 - 2 t ) ^ { 2 } = 5 t ^ { 2 } + 36$$

Solve for t

$$t=2\sqrt{311}-32\approx 3.270384177$$

$$t=-2\sqrt{311}-32\approx -67.270384177$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16-2t\right)^{2}.

$$256-64t+4t^{2}=5t^{2}+36$$

Subtract 5t^{2} from both sides.

$$256-64t+4t^{2}-5t^{2}=36$$

Combine 4t^{2} and -5t^{2} to get -t^{2}.

$$256-64t-t^{2}=36$$

Subtract 36 from both sides.

$$256-64t-t^{2}-36=0$$

Subtract 36 from 256 to get 220.

$$220-64t-t^{2}=0$$

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.

$$-t^{2}-64t+220=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -64 for b, and 220 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-64\right)±\sqrt{\left(-64\right)^{2}-4\left(-1\right)\times 220}}{2\left(-1\right)}$$

Square -64.

$$t=\frac{-\left(-64\right)±\sqrt{4096-4\left(-1\right)\times 220}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-64\right)±\sqrt{4096+4\times 220}}{2\left(-1\right)}$$

Multiply 4 times 220.

$$t=\frac{-\left(-64\right)±\sqrt{4096+880}}{2\left(-1\right)}$$

Add 4096 to 880.

$$t=\frac{-\left(-64\right)±\sqrt{4976}}{2\left(-1\right)}$$

Take the square root of 4976.

$$t=\frac{-\left(-64\right)±4\sqrt{311}}{2\left(-1\right)}$$

The opposite of -64 is 64.

$$t=\frac{64±4\sqrt{311}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{64±4\sqrt{311}}{-2}$$

Now solve the equation t=\frac{64±4\sqrt{311}}{-2} when ± is plus. Add 64 to 4\sqrt{311}.

$$t=\frac{4\sqrt{311}+64}{-2}$$

Divide 64+4\sqrt{311} by -2.

$$t=-2\sqrt{311}-32$$

Now solve the equation t=\frac{64±4\sqrt{311}}{-2} when ± is minus. Subtract 4\sqrt{311} from 64.

$$t=\frac{64-4\sqrt{311}}{-2}$$

Divide 64-4\sqrt{311} by -2.

$$t=2\sqrt{311}-32$$

The equation is now solved.

$$t=-2\sqrt{311}-32 t=2\sqrt{311}-32$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(16-2t\right)^{2}.

$$256-64t+4t^{2}=5t^{2}+36$$

Subtract 5t^{2} from both sides.

$$256-64t+4t^{2}-5t^{2}=36$$

Combine 4t^{2} and -5t^{2} to get -t^{2}.

$$256-64t-t^{2}=36$$

Subtract 256 from both sides.

$$-64t-t^{2}=36-256$$

Subtract 256 from 36 to get -220.

$$-64t-t^{2}=-220$$

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.

$$-t^{2}-64t=-220$$

Divide both sides by -1.

$$\frac{-t^{2}-64t}{-1}=-\frac{220}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\left(-\frac{64}{-1}\right)t=-\frac{220}{-1}$$

Divide -64 by -1.

$$t^{2}+64t=-\frac{220}{-1}$$

Divide -220 by -1.

$$t^{2}+64t=220$$

Divide 64, the coefficient of the x term, by 2 to get 32. Then add the square of 32 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+64t+32^{2}=220+32^{2}$$

Square 32.

$$t^{2}+64t+1024=220+1024$$

Add 220 to 1024.

$$t^{2}+64t+1024=1244$$

Factor t^{2}+64t+1024. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+32\right)^{2}=1244$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+32\right)^{2}}=\sqrt{1244}$$

Simplify.

$$t+32=2\sqrt{311} t+32=-2\sqrt{311}$$

Subtract 32 from both sides of the equation.

$$t=2\sqrt{311}-32 t=-2\sqrt{311}-32$$