Solve x^2+13x+36<0

$$x ^ { 2 } + 13 x + 36 < 0$$

Solve for x

$$x\in \left(-9,-4\right)$$

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.

$$x^{2}+13x+36=0$$

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

$$x=\frac{-13±\sqrt{13^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$x=\frac{-13±5}{2}$$

Solve the equation x=\frac{-13±5}{2} when ± is plus and when ± is minus.

$$x=-4 x=-9$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+4\right)\left(x+9\right)<0$$

For the product to be negative, x+4 and x+9 have to be of the opposite signs. Consider the case when x+4 is positive and x+9 is negative.

$$x+4>0 x+9<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+9 is positive and x+4 is negative.

$$x+9>0 x+4<0$$

The solution satisfying both inequalities is x\in \left(-9,-4\right).

$$x\in \left(-9,-4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-9,-4\right)$$