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 $ax^{2}+bx+c=0$.
$$x^{2}-7x+6=0$$
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 $6$ for $c$ in the quadratic formula.
Solve the equation $x=\frac{7±5}{2}$ when $±$ is plus and when $±$ is minus.
$$x=6$$ $$x=1$$
Rewrite the inequality by using the obtained solutions.
$$\left(x-6\right)\left(x-1\right)\leq 0$$
For the product to be $≤0$, one of the values $x-6$ and $x-1$ has to be $≥0$ and the other has to be $≤0$. Consider the case when $x-6\geq 0$ and $x-1\leq 0$.
$$x-6\geq 0$$ $$x-1\leq 0$$
This is false for any $x$.
$$x\in \emptyset$$
Consider the case when $x-6\leq 0$ and $x-1\geq 0$.
$$x-1\geq 0$$ $$x-6\leq 0$$
The solution satisfying both inequalities is $x\in \left[1,6\right]$.
$$x\in \begin{bmatrix}1,6\end{bmatrix}$$
The final solution is the union of the obtained solutions.
