Solve x^2+10x+24>0

$$x ^ { 2 } + 10 x + 24 > 0$$

Solve for x

$$x\in \left(-\infty,-6\right)\cup \left(-4,\infty\right)$$

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.

$$x^{2}+10x+24=0$$

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, 10 for b, and 24 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 24}}{2}$$

Do the calculations.

$$x=\frac{-10±2}{2}$$

Solve the equation x=\frac{-10±2}{2} when ± is plus and when ± is minus.

$$x=-4 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+4\right)\left(x+6\right)>0$$

For the product to be positive, x+4 and x+6 have to be both negative or both positive. Consider the case when x+4 and x+6 are both negative.

$$x+4<0 x+6<0$$

The solution satisfying both inequalities is x<-6.

$$x<-6$$

Consider the case when x+4 and x+6 are both positive.

$$x+6>0 x+4>0$$

The solution satisfying both inequalities is x>-4.

$$x>-4$$

The final solution is the union of the obtained solutions.

$$x<-6\text{; }x>-4$$