Solve for: x ^ 2-7 x+10 >= 0

Expression: $$x ^ { 2 } - 7 x + 10 \geq 0$$

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 $ax^{2}+bx+c=0$.

$$x^{2}-7x+10=0$$

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 $a$, $-7$ for $b$, and $10$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{7±3}{2}$$

Solve the equation $x=\frac{7±3}{2}$ when $±$ is plus and when $±$ is minus.

$$x=5$$ $$x=2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x-2\right)\geq 0$$

For the product to be $≥0$, $x-5$ and $x-2$ have to be both $≤0$ or both $≥0$. Consider the case when $x-5$ and $x-2$ are both $≤0$.

$$x-5\leq 0$$ $$x-2\leq 0$$

The solution satisfying both inequalities is $x\leq 2$.

$$x\leq 2$$

Consider the case when $x-5$ and $x-2$ are both $≥0$.

$$x-2\geq 0$$ $$x-5\geq 0$$

The solution satisfying both inequalities is $x\geq 5$.

$$x\geq 5$$

The final solution is the union of the obtained solutions.

$$x\leq 2\text{; }x\geq 5$$