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.