linear equation? Calculus review? I can't nail this problem... 
The given equation is either linear or equivalent to a linear equation. Solve the equation. (If there is no solution, enter NO SOLUTION. If all real numbers are solutions, enter REALS.)
$$\sqrt{3x} + \sqrt{12}= \frac{x+3}{\sqrt{3}}$$
Thanks for helping if you can!
 A: *

*Multiply each side of the equation by $\sqrt{3}$:
$3\sqrt{x}+6=x+3$

*Subtract $x+3$ from each side of the equation:
$-x+3\sqrt{x}+3=0$

*Replace $\sqrt{x}$ with $t$:
$-t^2+3t+3=0$

*Find the roots of the quadratic equation:
$t_{1,2}=\dfrac{-3\pm\sqrt{9+12}}{-2}\approx-0.79,3.79$

*Replace $t$ with $\sqrt{x}$:


*

*$\sqrt{x}\approx-0.79 \implies x\not\in\mathbb{R}$

*$\sqrt{x}\approx3.79 \implies x\approx14.37$





A: Since you say the equation is linear I'll assume it's
$$
\sqrt{3}x + \sqrt{12} = x + 3/\sqrt{3}
$$
This can be simplified to 
$$
\sqrt{3}x + 2\sqrt{3} = x + \sqrt{3}
$$
from which it follows that
$$
(\sqrt{3}-1)x =-\sqrt{3}
\\ x =-\frac{\sqrt{3}}{\sqrt{3}-1} \approx -2.366
$$

EDIT. The question has now changed, but since the equation is stated to be linear I suspect it should be
$$
\sqrt{3}x + \sqrt{12} = \frac{x + 3}{\sqrt{3}}
$$
to which the solution is
$$
x = -\tfrac{3}{2}
$$
A: Here's the idea:
Multiply both sides by $\sqrt{3}$ to get $$3\sqrt{x}+6=x+3$$
Now let $x=y^2$ and solve for $y$ (and then for $x$).
