Solve x^2-8x+15geq0

$$x ^ { 2 } - 8 x + 15 \geq 0$$

Solve for x

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

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.

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

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.

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

Do the calculations.

$$x=\frac{8±2}{2}$$

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

$$x=5 x=3$$

Rewrite the inequality by using the obtained solutions.

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

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

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

The solution satisfying both inequalities is x\leq 3.

$$x\leq 3$$

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

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

The solution satisfying both inequalities is x\geq 5.

$$x\geq 5$$

The final solution is the union of the obtained solutions.

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