Solve (t)=40t-5t^2

$$( t ) = 40 t - 5 t ^ { 2 }$$

Solve for t

$$t = \frac{39}{5} = 7\frac{4}{5} = 7.8$$

$$t=0$$

Steps Using Factoring

Subtract 40t from both sides.

$$t-40t=-5t^{2}$$

Combine t and -40t to get -39t.

$$-39t=-5t^{2}$$

Add 5t^{2} to both sides.

$$-39t+5t^{2}=0$$

Factor out t.

$$t\left(-39+5t\right)=0$$

To find equation solutions, solve t=0 and -39+5t=0.

$$t=0 t=\frac{39}{5}$$

Steps Using the Quadratic Formula

Subtract 40t from both sides.

$$t-40t=-5t^{2}$$

Combine t and -40t to get -39t.

$$-39t=-5t^{2}$$

Add 5t^{2} to both sides.

$$-39t+5t^{2}=0$$

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.

$$5t^{2}-39t=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -39 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}}}{2\times 5}$$

Take the square root of \left(-39\right)^{2}.

$$t=\frac{-\left(-39\right)±39}{2\times 5}$$

The opposite of -39 is 39.

$$t=\frac{39±39}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{39±39}{10}$$

Now solve the equation t=\frac{39±39}{10} when ± is plus. Add 39 to 39.

$$t=\frac{78}{10}$$

Reduce the fraction \frac{78}{10} to lowest terms by extracting and canceling out 2.

$$t=\frac{39}{5}$$

Now solve the equation t=\frac{39±39}{10} when ± is minus. Subtract 39 from 39.

$$t=\frac{0}{10}$$

Divide 0 by 10.

$$t=0$$

The equation is now solved.

$$t=\frac{39}{5} t=0$$

Steps for Completing the Square

Subtract 40t from both sides.

$$t-40t=-5t^{2}$$

Combine t and -40t to get -39t.

$$-39t=-5t^{2}$$

Add 5t^{2} to both sides.

$$-39t+5t^{2}=0$$

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.

$$5t^{2}-39t=0$$

Divide both sides by 5.

$$\frac{5t^{2}-39t}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$t^{2}-\frac{39}{5}t=\frac{0}{5}$$

Divide 0 by 5.

$$t^{2}-\frac{39}{5}t=0$$

Divide -\frac{39}{5}, the coefficient of the x term, by 2 to get -\frac{39}{10}. Then add the square of -\frac{39}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{39}{5}t+\left(-\frac{39}{10}\right)^{2}=\left(-\frac{39}{10}\right)^{2}$$

Square -\frac{39}{10} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{39}{5}t+\frac{1521}{100}=\frac{1521}{100}$$

Factor t^{2}-\frac{39}{5}t+\frac{1521}{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}.

$$\left(t-\frac{39}{10}\right)^{2}=\frac{1521}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{39}{10}\right)^{2}}=\sqrt{\frac{1521}{100}}$$

Simplify.

$$t-\frac{39}{10}=\frac{39}{10} t-\frac{39}{10}=-\frac{39}{10}$$

Add \frac{39}{10} to both sides of the equation.

$$t=\frac{39}{5} t=0$$