\(y^2+\left(2-\sqrt{3}\right)y-2\sqrt{3}=0\)
a = 1, b = 2 - \(\sqrt{3}\) , c = \(-2\sqrt{3}\)
Ta có: \(\Delta=b^2-2ac=\left(2-\sqrt{3}\right)^2-4.1.-2\sqrt{3}\)
\(\Delta=2^2-2.2\sqrt{3}.2+\left(\sqrt{3}\right)^2-4.1-2\sqrt{3}\)
Tiếp theo làm sao nữa v mn?