Question 672826
In the case of {{{2x^2-4x+4}}}, it is in the form {{{ax^2+bx+c}}}, so {{{a = 2}}}, {{{b = -4}}}, {{{c = 4}}}


Use the quadratic formula to solve for x


{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-4)+-sqrt((-4)^2-4(2)(4)))/(2(2))}}} Plug in {{{a = 2}}}, {{{b = -4}}}, {{{c = 4}}}


{{{x = (4+-sqrt(16-(32)))/(4)}}}


{{{x = (4+-sqrt(-16))/4}}}


{{{x = (4+sqrt(-16))/4}}} or {{{x = (4-sqrt(-16))/4}}}


{{{x = (4+4*i)/4}}} or {{{x = (4-4*i)/4}}}


{{{x = (4(1+i))/4}}} or {{{x = (4(1-i))/4}}}


{{{x = 1+i}}} or {{{x = 1-i}}}


So the two nonreal (or complex) solutions are {{{x = 1+i}}} or {{{x = 1-i}}}