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Tunajua kuwa $2^{2x+1} = 2^{2x} \cdot 2^1 = (2^x)^2 \cdot 2$.
Hivyo, mlinganyo $2^{2x+1} - 5 \cdot 2^x + 2 = 0$ unakuwa:

$$2(2^x)^2 - 5 \cdot 2^x + 2 = 0$$

Acha $y = 2^x$. Kwa kuwa $2^x$ daima ni chanya, $y > 0$.
Mlinganyo unakuwa:

$$2y^2 - 5y + 2 = 0$$

Tutatumia njia ya kuzidisha:

$$2y^2 - 4y - y + 2 = 0$$ $$2y(y - 2) - 1(y - 2) = 0$$ $$(2y - 1)(y - 2) = 0$$

Hii inatoa suluhisho mbili kwa $y$:

$$2y - 1 = 0 \implies 2y = 1 \implies y = \frac{1}{2}$$ $$y - 2 = 0 \implies y = 2$$

Thamani zote mbili za $y$ ni chanya, hivyo ni halali.

Kesi 1: $y = \frac{1}{2}$

$$2^x = \frac{1}{2}$$ $$2^x = 2^{-1}$$ $$x = -1$$

Kesi 2: $y = 2$

$$2^x = 2$$ $$2^x = 2^1$$ $$x = 1$$

Hivyo, maadili ya $x$ ni $-1$ na 1.

$$\text{x = -1, 1}$$