Expand the left side of the inequality.

$$x^2 - 11x + 30 \le 56$$

Subtract 56 from both sides:

$$x^2 - 11x + 30 - 56 \le 0$$ $$x^2 - 11x - 26 \le 0$$

We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=1$, $b=-11$, $c=-26$.
The discriminant is $\Delta = b^2 - 4ac = (-11)^2 - 4(1)(-26) = 121 + 104 = 225$.
The square root of the discriminant is $\sqrt{225} = 15$.

The roots are:

$$x_1 = \frac{-(-11) - 15}{2(1)} = \frac{11 - 15}{2} = \frac{-4}{2} = -2$$ $$x_2 = \frac{-(-11) + 15}{2(1)} = \frac{11 + 15}{2} = \frac{26}{2} = 13$$

So, the roots are $x = -2$ and $x = 13$.

Since the coefficient of $x^2$ is positive ($a=1$), the parabola opens upwards. This means the quadratic expression is less than or equal to zero between its roots, inclusive.

Therefore, the solution to the inequality is:

$$-2 \le x \le 13$$