Solve x^2+36x+240>0

$$x ^ { 2 } + 36 x + 240 > 0$$

Solve for x

$$x\in \left(-\infty,-2\sqrt{21}-18\right)\cup \left(2\sqrt{21}-18,\infty\right)$$

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.

$$x^{2}+36x+240=0$$

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, 36 for b, and 240 for c in the quadratic formula.

$$x=\frac{-36±\sqrt{36^{2}-4\times 1\times 240}}{2}$$

Do the calculations.

$$x=\frac{-36±4\sqrt{21}}{2}$$

Solve the equation x=\frac{-36±4\sqrt{21}}{2} when ± is plus and when ± is minus.

$$x=2\sqrt{21}-18 x=-2\sqrt{21}-18$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(2\sqrt{21}-18\right)\right)\left(x-\left(-2\sqrt{21}-18\right)\right)>0$$

For the product to be positive, x-\left(2\sqrt{21}-18\right) and x-\left(-2\sqrt{21}-18\right) have to be both negative or both positive. Consider the case when x-\left(2\sqrt{21}-18\right) and x-\left(-2\sqrt{21}-18\right) are both negative.

$$x-\left(2\sqrt{21}-18\right)<0 x-\left(-2\sqrt{21}-18\right)<0$$

The solution satisfying both inequalities is x<-2\sqrt{21}-18.

$$x<-2\sqrt{21}-18$$

Consider the case when x-\left(2\sqrt{21}-18\right) and x-\left(-2\sqrt{21}-18\right) are both positive.

$$x-\left(-2\sqrt{21}-18\right)>0 x-\left(2\sqrt{21}-18\right)>0$$

The solution satisfying both inequalities is x>2\sqrt{21}-18.

$$x>2\sqrt{21}-18$$

The final solution is the union of the obtained solutions.

$$x<-2\sqrt{21}-18\text{; }x>2\sqrt{21}-18$$