Solve m-6+15m^2>0

$$m - 6 + 15 m ^ { 2 } > 0$$

Solve for m

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

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.

$$m-6+15m^{2}=0$$

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, 1 for b, and -6 for c in the quadratic formula.

$$m=\frac{-1±\sqrt{1^{2}-4\times 15\left(-6\right)}}{2\times 15}$$

Do the calculations.

$$m=\frac{-1±19}{30}$$

Solve the equation m=\frac{-1±19}{30} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$15\left(m-\frac{3}{5}\right)\left(m+\frac{2}{3}\right)>0$$

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

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

The solution satisfying both inequalities is m<-\frac{2}{3}.

$$m<-\frac{2}{3}$$

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

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

The solution satisfying both inequalities is m>\frac{3}{5}.

$$m>\frac{3}{5}$$

The final solution is the union of the obtained solutions.

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