Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$ 
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$


Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-25\tag{2}$$
From $(1)$ and $(2)$, we have 
$$\alpha+\beta=\dfrac{(x-5)(x+5)}{x-5}=x+5\qquad(\therefore x\neq5)\tag{3}$$
From $(1)$ and $(3)$, we get
$$\alpha=\sqrt{3x^2-7x-30}=x\qquad\qquad\beta=\sqrt{2x^2-7x-5}=5$$
On solving any one of $\alpha=x$ or $\beta=5$, we get $x=6,-\dfrac{5}{2}$. 
On putting $x=-\dfrac{5}{2}$, $\alpha=x$ is not satisfied. Hence, $x=6$ is the only solution.
But as in $(3)$, we have ruled out $x=5$, as a solution then we can't put $x=5$ in the original equation. But, if we do put $x=5$ in the original equation we see that it is indeed satisfied. So, I tried a solution which also gives $x=5$ as a solution. So I picked the last solution from the $(2)$ without cancelling $(x-5)$.
$$(x-5)(\alpha+\beta)=x^2-25\implies (x-5)(\alpha+\beta-(x+5))=0$$.
Now, I am pretty much stuck here.
Edit-1:- I just saw now and feel pretty stupid about it.
If we proceed from the last step i.e. $(x-5)(\alpha+\beta-(x+5))=0$, we get 
$$(x-5)(\alpha+\beta-(x+5))=0\implies (x-5)=0\;\;\;\text{ or }\;\;\;\alpha+\beta=x+5$$
So we have the following equations to be solved:-
$$\alpha-\beta=x-5\\
\alpha+\beta=x+5\\
x-5=0$$
Which results in $x=5,6$.

Now, since I have put so much effort in posting this question, I might as well ask for better solutions if you can come up with one. And, please don't post a solution which includes a lot of squaring to result into a quartic equation.


Your $(2)$ is incorrect.  Instead, we should square both sides of $(1)$ to get
$$(\alpha-\beta)^2=(x-5)^2$$
$$\alpha^2-2\alpha\beta+\beta^2=(x-5)^2$$
$$2\alpha\beta=(x-5)^2-\alpha^2-\beta^2$$
Square both sides again
$$4\alpha^2\beta^2=\left[(x-5)^2-\alpha^2-\beta^2\right]^2$$
Now all radicals are removed and you should get something like two quartic polynomials.  Using the rational roots theorem, we get
$$x=5,6\tag{Click here}$$
And keeping in mind that we must have
$$3x^2-7x-30>0,\ 2x^2-7x-5>0$$
These are the only solutions.
