Solve -15m^2+84m+36>0

$$- 15 m ^ { 2 } + 84 m + 36 > 0$$

Solve for m

$$m\in \left(-\frac{2}{5},6\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -15m^{2}+84m+36 positive. Since -1 is negative, the inequality direction is changed.

$$15m^{2}-84m-36<0$$

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.

$$15m^{2}-84m-36=0$$

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 15 for a, -84 for b, and -36 for c in the quadratic formula.

$$m=\frac{-\left(-84\right)±\sqrt{\left(-84\right)^{2}-4\times 15\left(-36\right)}}{2\times 15}$$

Do the calculations.

$$m=\frac{84±96}{30}$$

Solve the equation m=\frac{84±96}{30} when ± is plus and when ± is minus.

$$m=6 m=-\frac{2}{5}$$

Rewrite the inequality by using the obtained solutions.

$$15\left(m-6\right)\left(m+\frac{2}{5}\right)<0$$

For the product to be negative, m-6 and m+\frac{2}{5} have to be of the opposite signs. Consider the case when m-6 is positive and m+\frac{2}{5} is negative.

$$m-6>0 m+\frac{2}{5}<0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m+\frac{2}{5} is positive and m-6 is negative.

$$m+\frac{2}{5}>0 m-6<0$$

The solution satisfying both inequalities is m\in \left(-\frac{2}{5},6\right).

$$m\in \left(-\frac{2}{5},6\right)$$

The final solution is the union of the obtained solutions.

$$m\in \left(-\frac{2}{5},6\right)$$