Rewrite the right side of the equation with the same base as the left side.

The equation is $3^{x^2+2} = 27^{2x-1}$.
Since $27 = 3^3$, we can substitute this into the equation.

$$3^{x^2+2} = (3^3)^{2x-1}$$

$$x^2 - 6x + 2 + 3 = 0$$ $$x^2 - 6x + 5 = 0$$

We need two numbers that multiply to 5 and add to -6. These numbers are -1 and -5.

$$(x-1)(x-5) = 0$$

$$x-1 = 0 \implies x = 1$$ $$x-5 = 0 \implies x = 5$$

The values of $x$ are 1 and 5.

$$x=1 \text{ or } x=5$$