Solve 16a^2-128a-144<0

$$16 a ^ { 2 } - 128 a - 144 < 0$$

Solve for a

$$a\in \left(-1,9\right)$$

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.

$$16a^{2}-128a-144=0$$

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, -128 for b, and -144 for c in the quadratic formula.

$$a=\frac{-\left(-128\right)±\sqrt{\left(-128\right)^{2}-4\times 16\left(-144\right)}}{2\times 16}$$

Do the calculations.

$$a=\frac{128±160}{32}$$

Solve the equation a=\frac{128±160}{32} when ± is plus and when ± is minus.

$$a=9 a=-1$$

Rewrite the inequality by using the obtained solutions.

$$16\left(a-9\right)\left(a+1\right)<0$$

For the product to be negative, a-9 and a+1 have to be of the opposite signs. Consider the case when a-9 is positive and a+1 is negative.

$$a-9>0 a+1<0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a+1 is positive and a-9 is negative.

$$a+1>0 a-9<0$$

The solution satisfying both inequalities is a\in \left(-1,9\right).

$$a\in \left(-1,9\right)$$

The final solution is the union of the obtained solutions.

$$a\in \left(-1,9\right)$$