I found something funny. Look at this particular equation
$$y = \sqrt{x^2 - x + 8}$$
Now suppose I want to find the domain for this graph. I know that the radical must be greater than or equal to zero (you can't take real solutions of negative radicands). Then all I do is set up an inequality to solve for the x values which are not possible by this graph
$$x^2 - x + 8 \ge 0$$
I solve by completing the square to get:
$$\left(x-\frac 1 2 \right) ^2 + \frac{31}{4} \ge 0$$
So far so good.
But wait. If you continue solving further, you end up having to square root a negative number...which in the real math world is not possible.
YET...if I put the initial graph function at the very top of this post on Desmos.com, I am able to get a nice graph.
WHAT IN THE WORLD?
Your key mis-step is assuming that finding $x$ satisfying the inequality
$$\left( x - \frac 1 2 \right)^2 + \frac{31}{4} \ge 0$$
means that you have to take the square root of a negative.
Think on a bit more of a basic level: no matter what you plug into $(x-1/2)^2$, what kind of numbers do you get? They're always nonnegative. This is seen in the graph of $f(x)=x^2$.
So, we know that the function you want to graph has domain $x$ which satisfy
$$\left( x - \frac 1 2 \right)^2 \ge - \frac{31}{4} $$
But we also know that, since we're squaring things, no matter what (real number) $x$ is,
$$\left( x - \frac 1 2 \right)^2 \ge 0 \ge - \frac{31}{4}$$
so, in fact, any real number $x$ satisfies our equation -- because the squaring essentially lets us "sandwich in" an inequality with $0$.