Question 44637
{{{y = x^2 - x + 5/4}}}
{{{y = x^2 - x + 1/4 + 4/4}}}
{{{y = (x - 1/2)^2 + 1}}}
let z = x - 1/2
{{{y = z^2 + 1}}}
to find roots use
{{{z = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
b = 0
a = 1
c = 1
{{{z =  (0 +- sqrt(0 - 4*1*1))/ (2*1)}}}
{{{z = -i}}}
{{{z = +i}}}
z = x - 1/2
{{{x - 1/2 = -i}}}
{{{x - 1/2 = +i}}}
{{{x = 1/2 - i}}} answer
{{{x = 1/2 + i}}} answer
roots are both imaginary
check by substitution in original equation