For a general quadratic #ax^2+bx+c#, you know that the solutions are given by #x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#.

First of all, we rewrite the equation as #-4x^2+4x-9=0#, which (if you prefer) can be also written as #4x^2-4x+9=0#

In this final case, #a=4#, #b=-4#, #c=9#. Substituing this values into the solving formula, we get

#x_{1,2}=\frac{4\pm\sqrt{16-4\cdot 4\cdot 9}}{2\cdot 4}#

The discriminant #b^2-4ac# is negative, and thus there are no real solutions.

If you're interested in complex solutions, you get

#x_{1,2}=\frac{4\pmi\sqrt{128}}{8}= \frac{4\pm8i\sqrt{2}}{8}#

which yelds

#x_1 = \frac{1}{2}+i\sqrt{2}#,

#x_2 = \frac{1}{2}-i\sqrt{2}#.