Solve k^2+20k-315>0

$$k ^ { 2 } + 20 k - 315 > 0$$

Solve for k

$$k\in \left(-\infty,-\sqrt{415}-10\right)\cup \left(\sqrt{415}-10,\infty\right)$$

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.

$$k^{2}+20k-315=0$$

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, 20 for b, and -315 for c in the quadratic formula.

$$k=\frac{-20±\sqrt{20^{2}-4\times 1\left(-315\right)}}{2}$$

Do the calculations.

$$k=\frac{-20±2\sqrt{415}}{2}$$

Solve the equation k=\frac{-20±2\sqrt{415}}{2} when ± is plus and when ± is minus.

$$k=\sqrt{415}-10 k=-\sqrt{415}-10$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-\left(\sqrt{415}-10\right)\right)\left(k-\left(-\sqrt{415}-10\right)\right)>0$$

For the product to be positive, k-\left(\sqrt{415}-10\right) and k-\left(-\sqrt{415}-10\right) have to be both negative or both positive. Consider the case when k-\left(\sqrt{415}-10\right) and k-\left(-\sqrt{415}-10\right) are both negative.

$$k-\left(\sqrt{415}-10\right)<0 k-\left(-\sqrt{415}-10\right)<0$$

The solution satisfying both inequalities is k<-\left(\sqrt{415}+10\right).

$$k<-\left(\sqrt{415}+10\right)$$

Consider the case when k-\left(\sqrt{415}-10\right) and k-\left(-\sqrt{415}-10\right) are both positive.

$$k-\left(-\sqrt{415}-10\right)>0 k-\left(\sqrt{415}-10\right)>0$$

The solution satisfying both inequalities is k>\sqrt{415}-10.

$$k>\sqrt{415}-10$$

The final solution is the union of the obtained solutions.

$$k<-\sqrt{415}-10\text{; }k>\sqrt{415}-10$$