the formula is (-b +/- sqrt(b^2 - 4ac)) /(2a) for an quadratic eqn ax^2 + bx + c = 0
if b^2 - a*c > 0, if will have two real roots, if it's equal to 0, you will have two equal real roots, if it's less than 0 you will have two conjugate(imaginary) roots
so for first case b^2 - 4ac = 36 - 36*2 = -36 so you have two imaginary roots as
-6 +/- sqrt(-36) / (2*9) = 1/3 +/- 1/3 i
second one, x1=x2 = 5
I hope you can solve the other two with the given formula