Solve 37/5t^2-44t+100=(10-2t)^2

$$\frac { 37 } { 5 } t ^ { 2 } - 44 t + 100 = ( 10 - 2 t ) ^ { 2 }$$

Solve for t

$$t = \frac{20}{17} = 1\frac{3}{17} \approx 1.176470588$$

$$t=0$$

Steps Using Factoring

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

$$\frac{37}{5}t^{2}-44t+100=100-40t+4t^{2}$$

Subtract 100 from both sides.

$$\frac{37}{5}t^{2}-44t+100-100=-40t+4t^{2}$$

Subtract 100 from 100 to get 0.

$$\frac{37}{5}t^{2}-44t=-40t+4t^{2}$$

Add 40t to both sides.

$$\frac{37}{5}t^{2}-44t+40t=4t^{2}$$

Combine -44t and 40t to get -4t.

$$\frac{37}{5}t^{2}-4t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$\frac{37}{5}t^{2}-4t-4t^{2}=0$$

Combine \frac{37}{5}t^{2} and -4t^{2} to get \frac{17}{5}t^{2}.

$$\frac{17}{5}t^{2}-4t=0$$

Factor out t.

$$t\left(\frac{17}{5}t-4\right)=0$$

To find equation solutions, solve t=0 and \frac{17t}{5}-4=0.

$$t=0 t=\frac{20}{17}$$

Steps Using the Quadratic Formula

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

$$\frac{37}{5}t^{2}-44t+100=100-40t+4t^{2}$$

Subtract 100 from both sides.

$$\frac{37}{5}t^{2}-44t+100-100=-40t+4t^{2}$$

Subtract 100 from 100 to get 0.

$$\frac{37}{5}t^{2}-44t=-40t+4t^{2}$$

Add 40t to both sides.

$$\frac{37}{5}t^{2}-44t+40t=4t^{2}$$

Combine -44t and 40t to get -4t.

$$\frac{37}{5}t^{2}-4t=4t^{2}$$

Subtract 4t^{2} from both sides.

$$\frac{37}{5}t^{2}-4t-4t^{2}=0$$

Combine \frac{37}{5}t^{2} and -4t^{2} to get \frac{17}{5}t^{2}.

$$\frac{17}{5}t^{2}-4t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{17}{5} for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times \frac{17}{5}}$$

Take the square root of \left(-4\right)^{2}.

$$t=\frac{-\left(-4\right)±4}{2\times \frac{17}{5}}$$

The opposite of -4 is 4.

$$t=\frac{4±4}{2\times \frac{17}{5}}$$

Multiply 2 times \frac{17}{5}.

$$t=\frac{4±4}{\frac{34}{5}}$$

Now solve the equation t=\frac{4±4}{\frac{34}{5}} when ± is plus. Add 4 to 4.

$$t=\frac{8}{\frac{34}{5}}$$

Divide 8 by \frac{34}{5} by multiplying 8 by the reciprocal of \frac{34}{5}.

$$t=\frac{20}{17}$$

Now solve the equation t=\frac{4±4}{\frac{34}{5}} when ± is minus. Subtract 4 from 4.

$$t=\frac{0}{\frac{34}{5}}$$

Divide 0 by \frac{34}{5} by multiplying 0 by the reciprocal of \frac{34}{5}.

$$t=0$$

The equation is now solved.

$$t=\frac{20}{17} t=0$$

Steps for Completing the Square

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

$$\frac{37}{5}t^{2}-44t+100=100-40t+4t^{2}$$

Add 40t to both sides.

$$\frac{37}{5}t^{2}-44t+100+40t=100+4t^{2}$$

Combine -44t and 40t to get -4t.

$$\frac{37}{5}t^{2}-4t+100=100+4t^{2}$$

Subtract 4t^{2} from both sides.

$$\frac{37}{5}t^{2}-4t+100-4t^{2}=100$$

Combine \frac{37}{5}t^{2} and -4t^{2} to get \frac{17}{5}t^{2}.

$$\frac{17}{5}t^{2}-4t+100=100$$

Subtract 100 from both sides.

$$\frac{17}{5}t^{2}-4t=100-100$$

Subtract 100 from 100 to get 0.

$$\frac{17}{5}t^{2}-4t=0$$

Divide both sides of the equation by \frac{17}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{17}{5}t^{2}-4t}{\frac{17}{5}}=\frac{0}{\frac{17}{5}}$$

Dividing by \frac{17}{5} undoes the multiplication by \frac{17}{5}.

$$t^{2}+\left(-\frac{4}{\frac{17}{5}}\right)t=\frac{0}{\frac{17}{5}}$$

Divide -4 by \frac{17}{5} by multiplying -4 by the reciprocal of \frac{17}{5}.

$$t^{2}-\frac{20}{17}t=\frac{0}{\frac{17}{5}}$$

Divide 0 by \frac{17}{5} by multiplying 0 by the reciprocal of \frac{17}{5}.

$$t^{2}-\frac{20}{17}t=0$$

Divide -\frac{20}{17}, the coefficient of the x term, by 2 to get -\frac{10}{17}. Then add the square of -\frac{10}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{20}{17}t+\left(-\frac{10}{17}\right)^{2}=\left(-\frac{10}{17}\right)^{2}$$

Square -\frac{10}{17} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{20}{17}t+\frac{100}{289}=\frac{100}{289}$$

Factor t^{2}-\frac{20}{17}t+\frac{100}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{10}{17}\right)^{2}=\frac{100}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{10}{17}\right)^{2}}=\sqrt{\frac{100}{289}}$$

Simplify.

$$t-\frac{10}{17}=\frac{10}{17} t-\frac{10}{17}=-\frac{10}{17}$$

Add \frac{10}{17} to both sides of the equation.

$$t=\frac{20}{17} t=0$$