help me please

Since we can't take the square root of a negative quantity, we need to find where -10x^2 - 11x + 6 = 0

Set  it  = 0  and multiply through by a -1....so we have

10x^2 + 11x - 6 = 0    factor, if possible

(5x - 2)(2x + 3) = 0

Setting each to 0, we have that x = 2/5 and  x = -3/2

Now, we need to find the intervals where -10x^2 - 11x + 6 < 0

This will occur at either  x < -3/2  or     -3/2 < x  < 2/5       or     x > 2/5

Letting x = 0 we see that -3/2 < x  < 2/5  "works"

If you pick any point in the other two intervals, the expression inside the radical will be < 0.

So, our answer is  -3/2 ≤ x ≤ 2/5

Here's a graph of the function that confirms our answer.....https://www.desmos.com/calculator/lhjgwiihyv

The domain of any square root function is D={xER} I dont know what you mean by endpoints when it never ends.

-.- Yea, dont listen to me I completely screwed this up

There has to be an end.

The bit under the square root must be greater or eqal to zero.

the easiest way to do questions like this is to let

$$y=-10x^2-11x+6$$

now you have

$$f(x)=\sqrt{y}$$

$$y\ge0\\$$

I just saw CPhill's answer pop up so there is no point in me continuing.

If you have questions just ask.  :)

so the endpoints are -3/2 and 2/5?

Yep.....