Solve x^2+6x-5>0

$${ x }^{ 2 } +6x-5 > 0$$

Solve for x

$$x\in \left(-\infty,-\sqrt{14}-3\right)\cup \left(\sqrt{14}-3,\infty\right)$$

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.

$$x^{2}+6x-5=0$$

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

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

Do the calculations.

$$x=\frac{-6±2\sqrt{14}}{2}$$

Solve the equation x=\frac{-6±2\sqrt{14}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{14}-3 x=-\sqrt{14}-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{14}-3\right)\right)\left(x-\left(-\sqrt{14}-3\right)\right)>0$$

For the product to be positive, x-\left(\sqrt{14}-3\right) and x-\left(-\sqrt{14}-3\right) have to be both negative or both positive. Consider the case when x-\left(\sqrt{14}-3\right) and x-\left(-\sqrt{14}-3\right) are both negative.

$$x-\left(\sqrt{14}-3\right)<0 x-\left(-\sqrt{14}-3\right)<0$$

The solution satisfying both inequalities is x<-\left(\sqrt{14}+3\right).

$$x<-\left(\sqrt{14}+3\right)$$

Consider the case when x-\left(\sqrt{14}-3\right) and x-\left(-\sqrt{14}-3\right) are both positive.

$$x-\left(-\sqrt{14}-3\right)>0 x-\left(\sqrt{14}-3\right)>0$$

The solution satisfying both inequalities is x>\sqrt{14}-3.

$$x>\sqrt{14}-3$$

The final solution is the union of the obtained solutions.

$$x<-\sqrt{14}-3\text{; }x>\sqrt{14}-3$$