Solve 3a^2-8a-16<0

$$3 a ^ { 2 } - 8 a - 16 < 0$$

Solve for a

$$a\in \left(-\frac{4}{3},4\right)$$

Steps Using the Quadratic Formula

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.

$$3a^{2}-8a-16=0$$

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 3 for a, -8 for b, and -16 for c in the quadratic formula.

$$a=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-16\right)}}{2\times 3}$$

Do the calculations.

$$a=\frac{8±16}{6}$$

Solve the equation a=\frac{8±16}{6} when ± is plus and when ± is minus.

$$a=4 a=-\frac{4}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(a-4\right)\left(a+\frac{4}{3}\right)<0$$

For the product to be negative, a-4 and a+\frac{4}{3} have to be of the opposite signs. Consider the case when a-4 is positive and a+\frac{4}{3} is negative.

$$a-4>0 a+\frac{4}{3}<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+\frac{4}{3} is positive and a-4 is negative.

$$a+\frac{4}{3}>0 a-4<0$$

The solution satisfying both inequalities is a\in \left(-\frac{4}{3},4\right).

$$a\in \left(-\frac{4}{3},4\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-\frac{4}{3},4\right)$$