Solve -4k^2+11k+3<0

$$- 4 k ^ { 2 } + 11 k + 3 < 0$$

Solve for k

$$k\in \left(-\infty,-\frac{1}{4}\right)\cup \left(3,\infty\right)$$

Steps Using the Quadratic Formula

Multiply the inequality by -1 to make the coefficient of the highest power in -4k^{2}+11k+3 positive. Since -1 is negative, the inequality direction is changed.

$$4k^{2}-11k-3>0$$

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.

$$4k^{2}-11k-3=0$$

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, -11 for b, and -3 for c in the quadratic formula.

$$k=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Do the calculations.

$$k=\frac{11±13}{8}$$

Solve the equation k=\frac{11±13}{8} when ± is plus and when ± is minus.

$$k=3 k=-\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(k-3\right)\left(k+\frac{1}{4}\right)>0$$

For the product to be positive, k-3 and k+\frac{1}{4} have to be both negative or both positive. Consider the case when k-3 and k+\frac{1}{4} are both negative.

$$k-3<0 k+\frac{1}{4}<0$$

The solution satisfying both inequalities is k<-\frac{1}{4}.

$$k<-\frac{1}{4}$$

Consider the case when k-3 and k+\frac{1}{4} are both positive.

$$k+\frac{1}{4}>0 k-3>0$$

The solution satisfying both inequalities is k>3.

$$k>3$$

The final solution is the union of the obtained solutions.

$$k<-\frac{1}{4}\text{; }k>3$$