Find angle!

The solutions of $x^2+\sqrt2x+1=0$ are $-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$ .

If an equation with real coefficients, and sides of a triangle should be real, has one of these solutions, it will have the other as well. So the equations will differ from $x^2+\sqrt2x+1=0$ only by a constant coefficient multiplying the left side. All such triangles will be similar, so for calculating the angle we can limit ourselves to the simplest form of the equations with $a=1, b=\sqrt{2}, c=1$ .

Since $a^2+c^2=b^2$ this forms an right isosceles triangle and $\angle C=45^\circ.$