Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$, I get $\sqrt{6-5}=\sqrt{1}=\pm 1$.
To me it seems that both values satisfy the original equation. I am using an online education system for my class called MyMathLab and the solution is only $x=1$. Why is that?
Thank you in advance.

The confusion you are having comes from the concept of finding the solutions to $z^2 = 36$. Definitely $z = \pm 6$ are solutions because $(-6)^2 = 36 = 6^2$, but this is not the same thing as $y = \sqrt{36}$. Otherwise, we get nonsense like $-6 = 6$ which isn't true.
If we plug the answer $x = -6$ back into the original equation, we have $$\sqrt{6 - 5(-6)} = \sqrt{36} = 6 = -6.$$ Again, we get nonsense of $-6 = 6$. Even it were positive and negative $6$, you'd have $-6 = \pm 6$. This is true for only one of the values, which shows it's not valid to assume that $\sqrt{36} = \pm 6$.
