Solve 5x^2+3x-8>0

$$5 { x }^{ 2 } +3x-8 > 0$$

Solve for x

$$x\in \left(-\infty,-\frac{8}{5}\right)\cup \left(1,\infty\right)$$

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.

$$5x^{2}+3x-8=0$$

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, 3 for b, and -8 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-8\right)}}{2\times 5}$$

Do the calculations.

$$x=\frac{-3±13}{10}$$

Solve the equation x=\frac{-3±13}{10} when ± is plus and when ± is minus.

$$x=1 x=-\frac{8}{5}$$

Rewrite the inequality by using the obtained solutions.

$$5\left(x-1\right)\left(x+\frac{8}{5}\right)>0$$

For the product to be positive, x-1 and x+\frac{8}{5} have to be both negative or both positive. Consider the case when x-1 and x+\frac{8}{5} are both negative.

$$x-1<0 x+\frac{8}{5}<0$$

The solution satisfying both inequalities is x<-\frac{8}{5}.

$$x<-\frac{8}{5}$$

Consider the case when x-1 and x+\frac{8}{5} are both positive.

$$x+\frac{8}{5}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<-\frac{8}{5}\text{; }x>1$$