Solve x^2+8x+15leq0

$$x ^ { 2 } + 8 x + 15 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$

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.

$$x^{2}+8x+15=0$$

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, 8 for b, and 15 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 15}}{2}$$

Do the calculations.

$$x=\frac{-8±2}{2}$$

Solve the equation x=\frac{-8±2}{2} when ± is plus and when ± is minus.

$$x=-3 x=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+3\right)\left(x+5\right)\leq 0$$

For the product to be ≤0, one of the values x+3 and x+5 has to be ≥0 and the other has to be ≤0. Consider the case when x+3\geq 0 and x+5\leq 0.

$$x+3\geq 0 x+5\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+3\leq 0 and x+5\geq 0.

$$x+5\geq 0 x+3\leq 0$$

The solution satisfying both inequalities is x\in \left[-5,-3\right].

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-5,-3\end{bmatrix}$$