$$x^{2}-7x+10=0$$
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.$$x\leq 2\text{; }x\geq 5$$