Solve 6x^2+11x-10<0

$$6 x ^ { 2 } + 11 x - 10 < 0$$

Solve for x

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

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.

$$6x^{2}+11x-10=0$$

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, 11 for b, and -10 for c in the quadratic formula.

$$x=\frac{-11±\sqrt{11^{2}-4\times 6\left(-10\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{-11±19}{12}$$

Solve the equation x=\frac{-11±19}{12} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

For the product to be negative, x-\frac{2}{3} and x+\frac{5}{2} have to be of the opposite signs. Consider the case when x-\frac{2}{3} is positive and x+\frac{5}{2} is negative.

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

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+\frac{5}{2} is positive and x-\frac{2}{3} is negative.

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

The solution satisfying both inequalities is x\in \left(-\frac{5}{2},\frac{2}{3}\right).

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

The final solution is the union of the obtained solutions.

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