Question 853665
Those two steps are good.  Now the right member becomes {{{3^(3^x-x^2)}}}.


Revising the equation then you have:
{{{(sqrt(3))^(x+2)=3^(3^x-x^2)}}}


You can make the left member be the base of 3, raised to a power:
{{{3^((1/2))^(x+2)=3^(3^x-x^2)}}}

{{{3^((x+2)/2)=3^(3^x-x^2)}}}----rendering cuts off the top; 3^((x+2)/2)=3^(3^x-x^2)


The left member is 3 raised to a power; the right member is 3 also raised to a power.  Those powers, exponents, are equal.


{{{(x+2)/2=3^x-x^2}}}
Still not easy to solve.
{{{(x+2)/2+x^2=3^x}}}
{{{x^2+(1/2)x+1=3^x}}}-----No ordinary algebra steps for this.  Left and right members are separate functions which have one or more points (maybe) of intersection.