Solve -6x^2+13x+5<0

$$-6 { x }^{ 2 } +13x+5 < 0$$

Solve for x

$$x\in \left(-\infty,-\frac{1}{3}\right)\cup \left(\frac{5}{2},\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -6x^{2}+13x+5 positive. Since -1 is negative, the inequality direction is changed.

$$6x^{2}-13x-5>0$$

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.

$$6x^{2}-13x-5=0$$

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 6 for a, -13 for b, and -5 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{13±17}{12}$$

Solve the equation x=\frac{13±17}{12} when ± is plus and when ± is minus.

$$x=\frac{5}{2} x=-\frac{1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(x-\frac{5}{2}\right)\left(x+\frac{1}{3}\right)>0$$

For the product to be positive, x-\frac{5}{2} and x+\frac{1}{3} have to be both negative or both positive. Consider the case when x-\frac{5}{2} and x+\frac{1}{3} are both negative.

$$x-\frac{5}{2}<0 x+\frac{1}{3}<0$$

The solution satisfying both inequalities is x<-\frac{1}{3}.

$$x<-\frac{1}{3}$$

Consider the case when x-\frac{5}{2} and x+\frac{1}{3} are both positive.

$$x+\frac{1}{3}>0 x-\frac{5}{2}>0$$

The solution satisfying both inequalities is x>\frac{5}{2}.

$$x>\frac{5}{2}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{3}\text{; }x>\frac{5}{2}$$