Solve 64a^2+48a-36>0

$$64 a ^ { 2 } + 48 a - 36 > 0$$

Solve for a

$$a\in \left(-\infty,\frac{-3\sqrt{5}-3}{8}\right)\cup \left(\frac{3\sqrt{5}-3}{8},\infty\right)$$

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.

$$64a^{2}+48a-36=0$$

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 64 for a, 48 for b, and -36 for c in the quadratic formula.

$$a=\frac{-48±\sqrt{48^{2}-4\times 64\left(-36\right)}}{2\times 64}$$

Do the calculations.

$$a=\frac{-48±48\sqrt{5}}{128}$$

Solve the equation a=\frac{-48±48\sqrt{5}}{128} when ± is plus and when ± is minus.

$$a=\frac{3\sqrt{5}-3}{8} a=\frac{-3\sqrt{5}-3}{8}$$

Rewrite the inequality by using the obtained solutions.

$$64\left(a-\frac{3\sqrt{5}-3}{8}\right)\left(a-\frac{-3\sqrt{5}-3}{8}\right)>0$$

For the product to be positive, a-\frac{3\sqrt{5}-3}{8} and a-\frac{-3\sqrt{5}-3}{8} have to be both negative or both positive. Consider the case when a-\frac{3\sqrt{5}-3}{8} and a-\frac{-3\sqrt{5}-3}{8} are both negative.

$$a-\frac{3\sqrt{5}-3}{8}<0 a-\frac{-3\sqrt{5}-3}{8}<0$$

The solution satisfying both inequalities is a<\frac{-3\sqrt{5}-3}{8}.

$$a<\frac{-3\sqrt{5}-3}{8}$$

Consider the case when a-\frac{3\sqrt{5}-3}{8} and a-\frac{-3\sqrt{5}-3}{8} are both positive.

$$a-\frac{-3\sqrt{5}-3}{8}>0 a-\frac{3\sqrt{5}-3}{8}>0$$

The solution satisfying both inequalities is a>\frac{3\sqrt{5}-3}{8}.

$$a>\frac{3\sqrt{5}-3}{8}$$

The final solution is the union of the obtained solutions.

$$a<\frac{-3\sqrt{5}-3}{8}\text{; }a>\frac{3\sqrt{5}-3}{8}$$