The slope of two-lines $6x^2-xy-2y^2=0$ differ by

Equation:

\(6x^2 - xy - 2y^2 = 0\)

Compare with: \( ax^2 + 2hxy + by^2 = 0 \)

\(a = 6\)

\(2h = -1 \;\;\Rightarrow\;\; h = -\tfrac{1}{2}\)

\(b = -2\)

Slopes satisfy:

\(bm^2 + 2hm + a = 0\)

\(-2m^2 - m + 6 = 0\)

\(\Rightarrow 2m^2 + m - 6 = 0\)

Solving quadratic:

\(m = \dfrac{-1 \pm \sqrt{49}}{4}\)

\(m = \dfrac{-1 \pm 7}{4}\)

Therefore:

\(m_1 = \tfrac{3}{2}, \quad m_2 = -2\)