Solve 8=-16t^2+8t+48

$$8 = - 16 t ^ { 2 } + 8 t + 48$$

Solve for t

$$t = \frac{\sqrt{41} + 1}{4} \approx 1.850781059$$

$$t=\frac{1-\sqrt{41}}{4}\approx -1.350781059$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-16t^{2}+8t+48=8$$

Subtract 8 from both sides.

$$-16t^{2}+8t+48-8=0$$

Subtract 8 from 48 to get 40.

$$-16t^{2}+8t+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 8 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-16\right)\times 40}}{2\left(-16\right)}$$

Square 8.

$$t=\frac{-8±\sqrt{64-4\left(-16\right)\times 40}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-8±\sqrt{64+64\times 40}}{2\left(-16\right)}$$

Multiply 64 times 40.

$$t=\frac{-8±\sqrt{64+2560}}{2\left(-16\right)}$$

Add 64 to 2560.

$$t=\frac{-8±\sqrt{2624}}{2\left(-16\right)}$$

Take the square root of 2624.

$$t=\frac{-8±8\sqrt{41}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-8±8\sqrt{41}}{-32}$$

Now solve the equation t=\frac{-8±8\sqrt{41}}{-32} when ± is plus. Add -8 to 8\sqrt{41}.

$$t=\frac{8\sqrt{41}-8}{-32}$$

Divide -8+8\sqrt{41} by -32.

$$t=\frac{1-\sqrt{41}}{4}$$

Now solve the equation t=\frac{-8±8\sqrt{41}}{-32} when ± is minus. Subtract 8\sqrt{41} from -8.

$$t=\frac{-8\sqrt{41}-8}{-32}$$

Divide -8-8\sqrt{41} by -32.

$$t=\frac{\sqrt{41}+1}{4}$$

The equation is now solved.

$$t=\frac{1-\sqrt{41}}{4} t=\frac{\sqrt{41}+1}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-16t^{2}+8t+48=8$$

Subtract 48 from both sides.

$$-16t^{2}+8t=8-48$$

Subtract 48 from 8 to get -40.

$$-16t^{2}+8t=-40$$

Divide both sides by -16.

$$\frac{-16t^{2}+8t}{-16}=-\frac{40}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{8}{-16}t=-\frac{40}{-16}$$

Reduce the fraction \frac{8}{-16} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{1}{2}t=-\frac{40}{-16}$$

Reduce the fraction \frac{-40}{-16} to lowest terms by extracting and canceling out 8.

$$t^{2}-\frac{1}{2}t=\frac{5}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{1}{2}t+\frac{1}{16}=\frac{5}{2}+\frac{1}{16}$$

Add \frac{5}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{1}{2}t+\frac{1}{16}=\frac{41}{16}$$

Factor t^{2}-\frac{1}{2}t+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{1}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$t-\frac{1}{4}=\frac{\sqrt{41}}{4} t-\frac{1}{4}=-\frac{\sqrt{41}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$t=\frac{\sqrt{41}+1}{4} t=\frac{1-\sqrt{41}}{4}$$