domain

To find the domain of the function $f(x) = \sqrt{-10x^2-11x+6+9x^2-5x}$, we need to determine the values of $x$ for which the expression under the square root is defined.

Simplify the expression under the square root:

$$-10x^2 - 11x + 6 + 9x^2 - 5x = -x^2 - 16x + 6.$$

The expression under the square root must not be negative, as the square root of a negative number is not a real number. Therefore, we need to find the values of $x$ for which $-x^2 - 16x + 6 \geq 0$.

To solve this inequality, we can factor the quadratic expression:

$$-x^2 - 16x + 6 = -(x^2 + 16x - 6).$$

Now, we want to find the values of $x$ that make the quadratic expression $x^2 + 16x - 6$ non-negative. We can do this by finding the roots of the quadratic equation $x^2 + 16x - 6 = 0$ and determining the intervals where the expression is positive or zero.

Factoring the quadratic equation $x^2 + 16x - 6 = 0$ is a bit tricky, so we can use the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

In this case, $a = 1$, $b = 16$, and $c = -6$, so the solutions are:

$$x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}.$$

Simplifying this gives:

$$x = -8 \pm \sqrt{130}.$$

Since the quadratic expression $x^2 + 16x - 6$ opens upwards (the coefficient of $x^2$ is positive), it is non-negative in the interval between its roots. Therefore, the values of $x$ that satisfy $-x^2 - 16x + 6 \geq 0$ are given by:

$$-8 - \sqrt{130} \leq x \leq -8 + \sqrt{130}.$$

In interval notation, the domain of the function $f(x)$ is:

$\boxed{[-8 - \sqrt{130}, -8 + \sqrt{130}]}.$