Solve -16x^2+16x+96>0

$$-16 { x }^{ 2 } +16x+96 > 0$$

Solve for x

$$x\in \left(-2,3\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -16x^{2}+16x+96 positive. Since -1 is negative, the inequality direction is changed.

$$16x^{2}-16x-96<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$16x^{2}-16x-96=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 16 for a, -16 for b, and -96 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{16±80}{32}$$

Solve the equation x=\frac{16±80}{32} when ± is plus and when ± is minus.

$$x=3 x=-2$$

Rewrite the inequality by using the obtained solutions.

$$16\left(x-3\right)\left(x+2\right)<0$$

For the product to be negative, x-3 and x+2 have to be of the opposite signs. Consider the case when x-3 is positive and x+2 is negative.

$$x-3>0 x+2<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+2 is positive and x-3 is negative.

$$x+2>0 x-3<0$$

The solution satisfying both inequalities is x\in \left(-2,3\right).

$$x\in \left(-2,3\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-2,3\right)$$