Question 38458
36x^2 - 18x + 9 = 0
It doesn't look factorable to me
divide through by 9
4x^2 -2x +1 = 0
dividing both sides by 4
x^2 -(1/2)x + 1/4 = 0
Using formula to find roots
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
b^2-4*a*c = 1/4 - 4*1*(1/4)
b^2-4*a*c = 1/4 - 4/4
b^2-4*a*c = -3/4
When the discriminant (under the square root sign) is negative,
the roots are imaginary
for roots, I get 
{{{(1 + sqrt(3)*i) /4}}}
and
{{{(1 - sqrt(3)*i) /4}}}