${2}^{2{x}^{2}-2x+2}={8}^{x}$
Write the expression in exponential form with the base of $2$${2}^{2{x}^{2}-2x+2}={2}^{3x}$
Since the bases are the same, set the exponents equal$2{x}^{2}-2x+2=3x$
Move the variable to the left-hand side and change its sign$2{x}^{2}-2x+2-3x=0$
Collect like terms$2{x}^{2}-5x+2=0$
Write $-5x$ as a difference$2{x}^{2}-x-4x+2=0$
Factor out $x$ from the expression$x \times \left( 2x-1 \right)-4x+2=0$
Factor out $-2$ from the expression$x \times \left( 2x-1 \right)-2\left( 2x-1 \right)=0$
Factor out $2x-1$ from the expression$\left( 2x-1 \right) \times \left( x-2 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }2x-1=0,\\x-2=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=\frac{ 1 }{ 2 },\\x-2=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=\frac{ 1 }{ 2 },\\x=2\end{array}$
The equation has $2$ solutions$\begin{align*}&\begin{array} { l }x_1=\frac{ 1 }{ 2 },& x_2=2\end{array} \\&\begin{array} { l }x_1=0.5,& x_2=2\end{array}\end{align*}$