Solve k^2-5k-6<0

$$k ^ { 2 } - 5 k - 6 < 0$$

Solve for k

$$k\in \left(-1,6\right)$$

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.

$$k^{2}-5k-6=0$$

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 1 for a, -5 for b, and -6 for c in the quadratic formula.

$$k=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-6\right)}}{2}$$

Do the calculations.

$$k=\frac{5±7}{2}$$

Solve the equation k=\frac{5±7}{2} when ± is plus and when ± is minus.

$$k=6 k=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-6\right)\left(k+1\right)<0$$

For the product to be negative, k-6 and k+1 have to be of the opposite signs. Consider the case when k-6 is positive and k+1 is negative.

$$k-6>0 k+1<0$$

This is false for any k.

$$k\in \emptyset$$

Consider the case when k+1 is positive and k-6 is negative.

$$k+1>0 k-6<0$$

The solution satisfying both inequalities is k\in \left(-1,6\right).

$$k\in \left(-1,6\right)$$

The final solution is the union of the obtained solutions.

$$k\in \left(-1,6\right)$$