Solving in-equation equations

Some equations look hard but contain a smaller equation inside them.
When one expression repeats, replace it with a single letter, solve the simpler equation that results, and then substitute back to find the original variable.

Take $2(x - 1)^2 - 5(x - 1) + 3 = 0$, where $(x - 1)$ repeats.
Let $m = x - 1$, turning it into the plain quadratic $2m^2 - 5m + 3 = 0$, which factors as $(2m - 3)(m - 1) = 0$, so $m = \frac{3}{2}$ or $m = 1$.
Now substitute back: $x - 1 = \frac{3}{2}$ gives $x = \frac{5}{2}$, and $x - 1 = 1$ gives $x = 2$.

The same trick works with exponents.
In $2^{2x} - 6(2^x) + 8 = 0$, notice that $2^{2x} = (2^x)^2$, so letting $m = 2^x$ gives $m^2 - 6m + 8 = 0$.
That factors as $(m - 2)(m - 4) = 0$, so $m = 2$ or $m = 4$; then $2^x = 2$ gives $x = 1$, and $2^x = 4$ gives $x = 2$.

So the method is always the same: spot the repeated expression, replace it with a single letter, solve the resulting quadratic, and substitute back.
Watch for solutions you must reject — for instance $2^x$ can never be zero or negative, so a negative value of $m$ would be thrown out.
The hard-looking equation is really just a simple one in disguise.

Solve $(x - 2)^2 + 3(x - 2) - 4 = 0$.
The expression $(x - 2)$ repeats, so let $m = x - 2$, giving $m^2 + 3m - 4 = 0$.
This factors as $(m + 4)(m - 1) = 0$, so $m = -4$ or $m = 1$.
Substituting back, $x - 2 = -4$ gives $x = -2$, and $x - 2 = 1$ gives $x = 3$.

Solve $2^{2x} - 5(2^x) + 4 = 0$.
Let $m = 2^x$, so $2^{2x} = m^2$ and the equation becomes $m^2 - 5m + 4 = 0$.
This factors as $(m - 1)(m - 4) = 0$, so $m = 1$ or $m = 4$.
Substituting back, $2^x = 1$ gives $x = 0$, and $2^x = 4$ gives $x = 2$.

Solve $3(2^{2x}) - 2(2^x) - 1 = 0$.
Let $m = 2^x$, so $2^{2x} = m^2$ and the equation becomes $3m^2 - 2m - 1 = 0$.
This factors as $(3m + 1)(m - 1) = 0$, so $m = -\frac{1}{3}$ or $m = 1$; but $2^x$ cannot be negative, so reject $m = -\frac{1}{3}$.
That leaves $2^x = 1$, which gives $x = 0$.