Solve x^2-15x+56geq0

$$x ^ { 2 } - 15 x + 56 \geq 0$$

Solve for x

$$x\in (-\infty,7]\cup [8,\infty)$$

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.

$$x^{2}-15x+56=0$$

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, -15 for b, and 56 for c in the quadratic formula.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 1\times 56}}{2}$$

Do the calculations.

$$x=\frac{15±1}{2}$$

Solve the equation x=\frac{15±1}{2} when ± is plus and when ± is minus.

$$x=8 x=7$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-8\right)\left(x-7\right)\geq 0$$

For the product to be ≥0, x-8 and x-7 have to be both ≤0 or both ≥0. Consider the case when x-8 and x-7 are both ≤0.

$$x-8\leq 0 x-7\leq 0$$

The solution satisfying both inequalities is x\leq 7.

$$x\leq 7$$

Consider the case when x-8 and x-7 are both ≥0.

$$x-7\geq 0 x-8\geq 0$$

The solution satisfying both inequalities is x\geq 8.

$$x\geq 8$$

The final solution is the union of the obtained solutions.

$$x\leq 7\text{; }x\geq 8$$