Solve x^2+9x+20<0

$${ x }^{ 2 } +9x+20 < 0$$

Solve for x

$$x\in \left(-5,-4\right)$$

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.

$$x^{2}+9x+20=0$$

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, 9 for b, and 20 for c in the quadratic formula.

$$x=\frac{-9±\sqrt{9^{2}-4\times 1\times 20}}{2}$$

Do the calculations.

$$x=\frac{-9±1}{2}$$

Solve the equation x=\frac{-9±1}{2} when ± is plus and when ± is minus.

$$x=-4 x=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+4\right)\left(x+5\right)<0$$

For the product to be negative, x+4 and x+5 have to be of the opposite signs. Consider the case when x+4 is positive and x+5 is negative.

$$x+4>0 x+5<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+5 is positive and x+4 is negative.

$$x+5>0 x+4<0$$

The solution satisfying both inequalities is x\in \left(-5,-4\right).

$$x\in \left(-5,-4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-5,-4\right)$$