8^{x-1 }= 9^{x+3 }Solve for x.

My attempt:

I'm guessing since they're both equal, if I multiply them it will just be the same as squaring one side.

We have (8^{x-1})(9^{x+3})=72^{x^2+2x-3 }

ANNNNND I don't know what to do next. I tried equating x^{2}+2x-3 = 2 since 72^{x^2+2x-3 }= a^{2}where a is the base. Solving for x^{2}+2x-5=0, I got:

$\displaystyle x = \ -1\pm\sqrt{6}$

How do I solve this? I need to find the base.