Solve 4x^2+9x-9<0

$$4 x ^ { 2 } + 9 x - 9 < 0$$

Solve for x

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

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.

$$4x^{2}+9x-9=0$$

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 4 for a, 9 for b, and -9 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 4\left(-9\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{-9±15}{8}$$

Solve the equation x=\frac{-9±15}{8} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any x.

$$x\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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