You need to use properties of exponentials for each term in equation such that:

`2^(2y-1) = 2^(2y)*(2^(-1))`

`2^(y-2) = 2^y*(2^(-2))`

You need to use negative power property such that:

`2^(2y)*(2^(-1)) = 2^(2y)/2`

`2^y*(2^(-2)) = (2^y)/(2^2)`

You need to write the equation such that:

`(2^(2y))/2 - 9(2^y)/(4) + 1 = 0`

You should come up with the substitution such that:

`2^y = x =gt 2^(2y) = x^2`

You need to write the equation in terms of x such that:

`x^2/2 - (9x)/4 + 1 = 0`

You need to bring the terms to a common denominator such that:

`2x^2 - 9x + 4 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (9 +- sqrt(81 - 32))/4`

`x_(1,2) = (9 +-sqrt49)/4`

`x_1 = (9+7)/4 =gt x_1 = 4`

`x_2 = (9-7)/4 =gt x_2 =1/2`

You need to solve for y the equations `2^y = 4` and `2^y = 1/2` such that:

`2^y = 4 =gt 2^y = 2^2 =gty = 2`

`2^y = 1/2 =gt 2^y = 2^(-1) =gt y = -1`

**Hence, evaluating the solutions to this equation yields y = -1 and y=2 (not 0).**

Notice that the terms in equation `x^2/2 - 9x/4 + 1 = 0` need to have the same denominator, hence `x^2/2` is multiplied by 2 and 1 is multiplied by 4, thus, the original term +1 becomes +4 after multiplication, hence +1 is not omitted.