Solve -4k^2+3k+16<0

$$- 4 k ^ { 2 } + 3 k + 16 < 0$$

Solve for k

$$k\in \left(-\infty,\frac{3-\sqrt{265}}{8}\right)\cup \left(\frac{\sqrt{265}+3}{8},\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -4k^{2}+3k+16 positive. Since -1 is negative, the inequality direction is changed.

$$4k^{2}-3k-16>0$$

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.

$$4k^{2}-3k-16=0$$

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 4 for a, -3 for b, and -16 for c in the quadratic formula.

$$k=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-16\right)}}{2\times 4}$$

Do the calculations.

$$k=\frac{3±\sqrt{265}}{8}$$

Solve the equation k=\frac{3±\sqrt{265}}{8} when ± is plus and when ± is minus.

$$k=\frac{\sqrt{265}+3}{8} k=\frac{3-\sqrt{265}}{8}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(k-\frac{\sqrt{265}+3}{8}\right)\left(k-\frac{3-\sqrt{265}}{8}\right)>0$$

For the product to be positive, k-\frac{\sqrt{265}+3}{8} and k-\frac{3-\sqrt{265}}{8} have to be both negative or both positive. Consider the case when k-\frac{\sqrt{265}+3}{8} and k-\frac{3-\sqrt{265}}{8} are both negative.

$$k-\frac{\sqrt{265}+3}{8}<0 k-\frac{3-\sqrt{265}}{8}<0$$

The solution satisfying both inequalities is k<\frac{3-\sqrt{265}}{8}.

$$k<\frac{3-\sqrt{265}}{8}$$

Consider the case when k-\frac{\sqrt{265}+3}{8} and k-\frac{3-\sqrt{265}}{8} are both positive.

$$k-\frac{3-\sqrt{265}}{8}>0 k-\frac{\sqrt{265}+3}{8}>0$$

The solution satisfying both inequalities is k>\frac{\sqrt{265}+3}{8}.

$$k>\frac{\sqrt{265}+3}{8}$$

The final solution is the union of the obtained solutions.

$$k<\frac{3-\sqrt{265}}{8}\text{; }k>\frac{\sqrt{265}+3}{8}$$