An algebraic exponent problem by Ad

By assuming 2^x as y,we get the equation as , 2y^2-9y+4=0. By factorising,we get the value of y as 1/2 ,4. Since y=2^x,the value of x is -1 ,2

$\begin{array}{l} {2^{(2x + 1)}} - 9 \times {2^x} + 4 = 0\\ \Leftrightarrow 2 \times {2^{2x}} - 9 \times {2^x} + 4 = 0\\ \Leftrightarrow 2 \times {\left( {{2^x}} \right)^2} - 9 \times {2^x} + 4 = 0\\ \Leftrightarrow {2^x} = \frac{{9 \pm \sqrt {{{( - 9)}^2} - 4 \times 2 \times 4} }}{{2 \times 2}}\\ \Leftrightarrow {2^x} = 4\;\; \vee \;\;{2^x} = \frac{1}{2}\\ \Leftrightarrow {2^x} = {2^2}\;\; \vee \;\;{2^x} = {2^{ - 1}}\\ \Leftrightarrow x = 2\;\; \vee \;\;x = - 1 \end{array}$