Solve 3k^2+8k-4<0

$$3 k ^ { 2 } + 8 k - 4 < 0$$

Solve for k

$$k\in \left(\frac{-2\sqrt{7}-4}{3},\frac{2\sqrt{7}-4}{3}\right)$$

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.

$$3k^{2}+8k-4=0$$

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 3 for a, 8 for b, and -4 for c in the quadratic formula.

$$k=\frac{-8±\sqrt{8^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Do the calculations.

$$k=\frac{-8±4\sqrt{7}}{6}$$

Solve the equation k=\frac{-8±4\sqrt{7}}{6} when ± is plus and when ± is minus.

$$k=\frac{2\sqrt{7}-4}{3} k=\frac{-2\sqrt{7}-4}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(k-\frac{2\sqrt{7}-4}{3}\right)\left(k-\frac{-2\sqrt{7}-4}{3}\right)<0$$

For the product to be negative, k-\frac{2\sqrt{7}-4}{3} and k-\frac{-2\sqrt{7}-4}{3} have to be of the opposite signs. Consider the case when k-\frac{2\sqrt{7}-4}{3} is positive and k-\frac{-2\sqrt{7}-4}{3} is negative.

$$k-\frac{2\sqrt{7}-4}{3}>0 k-\frac{-2\sqrt{7}-4}{3}<0$$

This is false for any k.

$$k\in \emptyset$$

Consider the case when k-\frac{-2\sqrt{7}-4}{3} is positive and k-\frac{2\sqrt{7}-4}{3} is negative.

$$k-\frac{-2\sqrt{7}-4}{3}>0 k-\frac{2\sqrt{7}-4}{3}<0$$

The solution satisfying both inequalities is k\in \left(\frac{-2\sqrt{7}-4}{3},\frac{2\sqrt{7}-4}{3}\right).

$$k\in \left(\frac{-2\sqrt{7}-4}{3},\frac{2\sqrt{7}-4}{3}\right)$$

The final solution is the union of the obtained solutions.

$$k\in \left(\frac{-2\sqrt{7}-4}{3},\frac{2\sqrt{7}-4}{3}\right)$$