Solve x^2+2x-15geq0

$$x ^ { 2 } + 2 x - 15 \geq 0$$

Solve for x

$$x\in (-\infty,-5]\cup [3,\infty)$$

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.

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

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, 2 for b, and -15 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-15\right)}}{2}$$

Do the calculations.

$$x=\frac{-2±8}{2}$$

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

$$x=3 x=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-3\right)\left(x+5\right)\geq 0$$

For the product to be ≥0, x-3 and x+5 have to be both ≤0 or both ≥0. Consider the case when x-3 and x+5 are both ≤0.

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

The solution satisfying both inequalities is x\leq -5.

$$x\leq -5$$

Consider the case when x-3 and x+5 are both ≥0.

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

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq -5\text{; }x\geq 3$$