How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms?

You and every other answer thus far have talked about starting by squaring both sides, which works, but there is a slightly easier path to take. Rewrite your equation so that there is only one square root on each side, for example: $$4\sqrt{x-3}=3+\sqrt{6x-17}$$ Now, when you square both sides, you'll still have another square root to deal with, but it's not the square root of a product, like the $\sqrt{(x-3)(6x-17)}$ you would have had: $$16(x-3)=9+6\sqrt{6x-17}+(6x-17)$$ Expand, collect like terms, and rearrange to get the square root by itself: $$5x-20=3\sqrt{6x-17}$$ Square both sides again: $$25x^2-200x+400=9(6x-17)$$ $$25x^2-254x+553=0$$ Solving this gives $$x=7\text{ or }x=\frac{79}{25}$$ but $x=\frac{79}{25}$ doesn't work in the original equation, so $x=7$ is the only solution to the original equation.

If you want less square roots, you may try replacing $x=t^2+3$ to simplify the equation to get $$4t-\sqrt{6t^2+1}=3$$then$$(4t-3)^2=6t^2+1$$and finally$$5t^2-12t+4=0$$which gives you $t=2$ (i.e. $x=7$) or $t=2/5$ (i.e. $x=79/25$).

You can square the equation and you end up with $$16(x-3)+(6x-17)-9=8\sqrt{(x-3)(6x-17)}$$ i.e. $$22x-74=8\sqrt{(x-3)(6x-17)} $$ and square again to obtain $$484x^2-44\cdot 74 x+74^2=64\cdot(x-3)(6x-17)$$ which is a second degree equation in $x$ you know how to solve! This yields at most $2$ solutions to your problem, and you can check by hand if both solutions work or if one doesn't work for instance because the term inside the square root is negative...