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