Solve 5x^2-18x-8<0

$$5 { x }^{ 2 } -18x-8 < 0$$

Solve for x

$$x\in \left(-\frac{2}{5},4\right)$$

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.

$$5x^{2}-18x-8=0$$

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 5 for a, -18 for b, and -8 for c in the quadratic formula.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 5\left(-8\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{18±22}{10}$$

Solve the equation x=\frac{18±22}{10} when ± is plus and when ± is minus.

$$x=4 x=-\frac{2}{5}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-4\right)\left(x+\frac{2}{5}\right)<0$$

For the product to be negative, x-4 and x+\frac{2}{5} have to be of the opposite signs. Consider the case when x-4 is positive and x+\frac{2}{5} is negative.

$$x-4>0 x+\frac{2}{5}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{2}{5} is positive and x-4 is negative.

$$x+\frac{2}{5}>0 x-4<0$$

The solution satisfying both inequalities is x\in \left(-\frac{2}{5},4\right).

$$x\in \left(-\frac{2}{5},4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{2}{5},4\right)$$