When $$a \ne 0$$, there are two solutions to $$ax^2 + bx + c = 0$$ and they are $x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$
non ho capito $$x^2-3x+1$$ oppure $x^2-3x+1$ $\dfrac{3}{\sqrt{12}}$

x2+y2+ax+bx+c=0
x1,2=3
x1,2=-b±√b2-4ac2provaciao