Solve 0=-16t^2+80t+96

$$0 = - 16 t ^ { 2 } + 80 t + 96$$

Solve for t

$$t=-1$$

$$t=6$$

Steps Using Factoring By Grouping

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

$$-16t^{2}+80t+96=0$$

Divide both sides by 16.

$$-t^{2}+5t+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt+6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=6 b=-1$$

Rewrite -t^{2}+5t+6 as \left(-t^{2}+6t\right)+\left(-t+6\right).

$$\left(-t^{2}+6t\right)+\left(-t+6\right)$$

Factor out -t in the first and -1 in the second group.

$$-t\left(t-6\right)-\left(t-6\right)$$

Factor out common term t-6 by using distributive property.

$$\left(t-6\right)\left(-t-1\right)$$

To find equation solutions, solve t-6=0 and -t-1=0.

$$t=6 t=-1$$

Steps Using the Quadratic Formula

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

$$-16t^{2}+80t+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 80 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-80±\sqrt{80^{2}-4\left(-16\right)\times 96}}{2\left(-16\right)}$$

Square 80.

$$t=\frac{-80±\sqrt{6400-4\left(-16\right)\times 96}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-80±\sqrt{6400+64\times 96}}{2\left(-16\right)}$$

Multiply 64 times 96.

$$t=\frac{-80±\sqrt{6400+6144}}{2\left(-16\right)}$$

Add 6400 to 6144.

$$t=\frac{-80±\sqrt{12544}}{2\left(-16\right)}$$

Take the square root of 12544.

$$t=\frac{-80±112}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-80±112}{-32}$$

Now solve the equation t=\frac{-80±112}{-32} when ± is plus. Add -80 to 112.

$$t=\frac{32}{-32}$$

Divide 32 by -32.

$$t=-1$$

Now solve the equation t=\frac{-80±112}{-32} when ± is minus. Subtract 112 from -80.

$$t=-\frac{192}{-32}$$

Divide -192 by -32.

$$t=6$$

The equation is now solved.

$$t=-1 t=6$$

Steps for Completing the Square

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

$$-16t^{2}+80t+96=0$$

Subtract 96 from both sides. Anything subtracted from zero gives its negation.

$$-16t^{2}+80t=-96$$

Divide both sides by -16.

$$\frac{-16t^{2}+80t}{-16}=-\frac{96}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{80}{-16}t=-\frac{96}{-16}$$

Divide 80 by -16.

$$t^{2}-5t=-\frac{96}{-16}$$

Divide -96 by -16.

$$t^{2}-5t=6$$

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

$$t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-5t+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{49}{4}$$

Factor t^{2}-5t+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$t-\frac{5}{2}=\frac{7}{2} t-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$t=6 t=-1$$