3^[(sqrt x) -1]+3^[2-(sqrt x)]=12
This forum is for introductions. The forums most appropriate for this topic is either algebra or pre-calculus.
Anyway, I think I would rewrite the equation as follows:
$\displaystyle 3^{\sqrt{x}-1}+3^{2-\sqrt{x}}-12=0$
Multiply through by $\displaystyle 3^{\sqrt{x}+1}=3\cdot3^{\sqrt{x}}$:
$\displaystyle \left(3^{\sqrt{x}} \right)^2-36\left(3^{\sqrt{x}} \right)+27=0$
Now you have a quadratic in $\displaystyle 3^{\sqrt{x}$.