Solve (t)=-5t^2+20

$$( t ) = - 5 t ^ { 2 } + 20$$

Solve for t

$$t = \frac{\sqrt{401} - 1}{10} \approx 1.902498439$$

$$t=\frac{-\sqrt{401}-1}{10}\approx -2.102498439$$

Steps Using the Quadratic Formula

Add 5t^{2} to both sides.

$$t+5t^{2}=20$$

Subtract 20 from both sides.

$$t+5t^{2}-20=0$$

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.

$$5t^{2}+t-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 1 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\times 5\left(-20\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-1±\sqrt{1-20\left(-20\right)}}{2\times 5}$$

Multiply -20 times -20.

$$t=\frac{-1±\sqrt{1+400}}{2\times 5}$$

Add 1 to 400.

$$t=\frac{-1±\sqrt{401}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-1±\sqrt{401}}{10}$$

Now solve the equation t=\frac{-1±\sqrt{401}}{10} when ± is plus. Add -1 to \sqrt{401}.

$$t=\frac{\sqrt{401}-1}{10}$$

Now solve the equation t=\frac{-1±\sqrt{401}}{10} when ± is minus. Subtract \sqrt{401} from -1.

$$t=\frac{-\sqrt{401}-1}{10}$$

The equation is now solved.

$$t=\frac{\sqrt{401}-1}{10} t=\frac{-\sqrt{401}-1}{10}$$

Steps for Completing the Square

Add 5t^{2} to both sides.

$$t+5t^{2}=20$$

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.

$$5t^{2}+t=20$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

$$t^{2}+\frac{1}{5}t=\frac{20}{5}$$

Divide 20 by 5.

$$t^{2}+\frac{1}{5}t=4$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{1}{5}t+\left(\frac{1}{10}\right)^{2}=4+\left(\frac{1}{10}\right)^{2}$$

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{1}{5}t+\frac{1}{100}=4+\frac{1}{100}$$

Add 4 to \frac{1}{100}.

$$t^{2}+\frac{1}{5}t+\frac{1}{100}=\frac{401}{100}$$

Factor t^{2}+\frac{1}{5}t+\frac{1}{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}.

$$\left(t+\frac{1}{10}\right)^{2}=\frac{401}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{10}\right)^{2}}=\sqrt{\frac{401}{100}}$$

Simplify.

$$t+\frac{1}{10}=\frac{\sqrt{401}}{10} t+\frac{1}{10}=-\frac{\sqrt{401}}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$t=\frac{\sqrt{401}-1}{10} t=\frac{-\sqrt{401}-1}{10}$$