Solve 132x^2+84x+9<0

$$132 { x }^{ 2 } +84x+9 < 0$$

Solve for x

$$x\in \left(-\frac{1}{2},-\frac{3}{22}\right)$$

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.

$$132x^{2}+84x+9=0$$

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 132 for a, 84 for b, and 9 for c in the quadratic formula.

$$x=\frac{-84±\sqrt{84^{2}-4\times 132\times 9}}{2\times 132}$$

Do the calculations.

$$x=\frac{-84±48}{264}$$

Solve the equation x=\frac{-84±48}{264} when ± is plus and when ± is minus.

$$x=-\frac{3}{22} x=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$132\left(x+\frac{3}{22}\right)\left(x+\frac{1}{2}\right)<0$$

For the product to be negative, x+\frac{3}{22} and x+\frac{1}{2} have to be of the opposite signs. Consider the case when x+\frac{3}{22} is positive and x+\frac{1}{2} is negative.

$$x+\frac{3}{22}>0 x+\frac{1}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{1}{2} is positive and x+\frac{3}{22} is negative.

$$x+\frac{1}{2}>0 x+\frac{3}{22}<0$$

The solution satisfying both inequalities is x\in \left(-\frac{1}{2},-\frac{3}{22}\right).

$$x\in \left(-\frac{1}{2},-\frac{3}{22}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{1}{2},-\frac{3}{22}\right)$$