Solve 15-m-16m^2<0

$$15 - m - 16 m ^ { 2 } < 0$$

Solve for m

$$m\in \left(-\infty,-1\right)\cup \left(\frac{15}{16},\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in 15-m-16m^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-15+m+16m^{2}>0$$

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.

$$-15+m+16m^{2}=0$$

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 16 for a, 1 for b, and -15 for c in the quadratic formula.

$$m=\frac{-1±\sqrt{1^{2}-4\times 16\left(-15\right)}}{2\times 16}$$

Do the calculations.

$$m=\frac{-1±31}{32}$$

Solve the equation m=\frac{-1±31}{32} when ± is plus and when ± is minus.

$$m=\frac{15}{16} m=-1$$

Rewrite the inequality by using the obtained solutions.

$$16\left(m-\frac{15}{16}\right)\left(m+1\right)>0$$

For the product to be positive, m-\frac{15}{16} and m+1 have to be both negative or both positive. Consider the case when m-\frac{15}{16} and m+1 are both negative.

$$m-\frac{15}{16}<0 m+1<0$$

The solution satisfying both inequalities is m<-1.

$$m<-1$$

Consider the case when m-\frac{15}{16} and m+1 are both positive.

$$m+1>0 m-\frac{15}{16}>0$$

The solution satisfying both inequalities is m>\frac{15}{16}.

$$m>\frac{15}{16}$$

The final solution is the union of the obtained solutions.

$$m<-1\text{; }m>\frac{15}{16}$$