Solve -16t^2+40t+96=0

$$-16 { t }^{ 2 } +40t+96=0$$

Solve for t

$$t=4$$

$$t = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Factoring By Grouping

Divide both sides by 8.

$$-2t^{2}+5t+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2t^{2}+at+bt+12. To find a and b, set up a system to be solved.

$$a+b=5 ab=-2\times 12=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=8 b=-3$$

Rewrite -2t^{2}+5t+12 as \left(-2t^{2}+8t\right)+\left(-3t+12\right).

$$\left(-2t^{2}+8t\right)+\left(-3t+12\right)$$

Factor out 2t in the first and 3 in the second group.

$$2t\left(-t+4\right)+3\left(-t+4\right)$$

Factor out common term -t+4 by using distributive property.

$$\left(-t+4\right)\left(2t+3\right)$$

To find equation solutions, solve -t+4=0 and 2t+3=0.

$$t=4 t=-\frac{3}{2}$$

Steps Using the Quadratic Formula

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.

$$-16t^{2}+40t+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 40 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-40±\sqrt{40^{2}-4\left(-16\right)\times 96}}{2\left(-16\right)}$$

Square 40.

$$t=\frac{-40±\sqrt{1600-4\left(-16\right)\times 96}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-40±\sqrt{1600+64\times 96}}{2\left(-16\right)}$$

Multiply 64 times 96.

$$t=\frac{-40±\sqrt{1600+6144}}{2\left(-16\right)}$$

Add 1600 to 6144.

$$t=\frac{-40±\sqrt{7744}}{2\left(-16\right)}$$

Take the square root of 7744.

$$t=\frac{-40±88}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-40±88}{-32}$$

Now solve the equation t=\frac{-40±88}{-32} when ± is plus. Add -40 to 88.

$$t=\frac{48}{-32}$$

Reduce the fraction \frac{48}{-32} to lowest terms by extracting and canceling out 16.

$$t=-\frac{3}{2}$$

Now solve the equation t=\frac{-40±88}{-32} when ± is minus. Subtract 88 from -40.

$$t=-\frac{128}{-32}$$

Divide -128 by -32.

$$t=4$$

The equation is now solved.

$$t=-\frac{3}{2} t=4$$

Steps for Completing the Square

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.

$$-16t^{2}+40t+96=0$$

Subtract 96 from both sides of the equation.

$$-16t^{2}+40t+96-96=-96$$

Subtracting 96 from itself leaves 0.

$$-16t^{2}+40t=-96$$

Divide both sides by -16.

$$\frac{-16t^{2}+40t}{-16}=-\frac{96}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$t^{2}+\frac{40}{-16}t=-\frac{96}{-16}$$

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

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

Divide -96 by -16.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{5}{2}t+\left(-\frac{5}{4}\right)^{2}=6+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{5}{2}t+\frac{25}{16}=6+\frac{25}{16}$$

Add 6 to \frac{25}{16}.

$$t^{2}-\frac{5}{2}t+\frac{25}{16}=\frac{121}{16}$$

Factor t^{2}-\frac{5}{2}t+\frac{25}{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}.

$$\left(t-\frac{5}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$t-\frac{5}{4}=\frac{11}{4} t-\frac{5}{4}=-\frac{11}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$t=4 t=-\frac{3}{2}$$