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}.
