We begin with #y=-5x^2+7x+2(2x-1)^2#. In order to use the quadratic formula, we need to get this equation into standard form, which means that the format needs to look like this: #ax^2+bx+c#.
So, in order to change the form of our current equation, we first need to expand everything that we can. Let's start with #(2x-1)^2#. This becomes #(2x-1)(2x-1)#, which is #4x^2-4x+1#. Now we have #-5x^2+7x+2(4x^2-4x+1)#/ By multiplying out the #2#, we arrive at #-5x^2+7x+8x^2-8x+2#.
From here, we just need to combine like terms. #-5x^2+8x^2# gives us #3x^2#, while #7x-8x# equals #-x#. The #2# is left alone.
So now we have the following: #y=3x^2-x+2#, which is in standard form, meaning that we are good to go for using the quadratic formula!
#(-b+-sqrt(b^2-4*a*c))/(2*a)#
#color(blue)(a=3)#
#color(green)(b=-1)#
#color(red)(c=2)#
We just need to substitute the #a#s and #b#s and #c#s for their numerical value.
The formula now becomes
#(-(color(green)(-1))+-sqrt((color(green)(-1))^2-4*(color(blue)(3))*(color(red)(2))))/(2(color(blue)(3)))#
Simplifying that we have #(1+-sqrt(1-24))/6# or #(1+-isqrt(23))/6#.
That is the imaginary root for #y=-5x^2+7x+2(2x-1)^2#. Nice job.
