We need to solve the equation:
[tex]2\nu²-2\nu+60=(\nu-6)²[/tex]
Expanding the right side, we obtain:
[tex]2\nu²-2\nu+60=\nu²-12\nu+36[/tex]
And applying the same operations on both sides, we obtain:
[tex]\begin{gathered} 2\nu²-2\nu+60-(\nu²-12\nu+36)=\nu²-12\nu+36-(\nu²-12\nu+36) \\ \\ 2\nu²-2\nu+60-\nu²+12\nu-36=0 \\ \\ \nu²+10\nu+24=0 \end{gathered}[/tex]
Now, using the quadratic formula, we obtain:
[tex]\begin{gathered} \nu=\frac{-10\pm\sqrt{10²-4(1)(24)}}{2(1)} \\ \\ \nu=\frac{-10\pm\sqrt{4}}{2} \\ \\ \nu=\frac{-10\pm2}{2} \\ \\ \nu=-5\pm1 \\ \\ \nu_1=-5-1=-6 \\ \\ \nu_2=-5+1=-4 \end{gathered}[/tex]
Answer: ν = -6, -4
