In order to solve this quadratic equation by factoring, we can do the following steps:
[tex]\begin{gathered} x^2-3x-10=0\\ \\ x^2-3x-5\cdot2=0\\ \\ x^2+2x-5x-5\cdot2=0\\ \\ x(x+2)-5(x+2)=0\\ \\ (x-5)(x+2)=0\\ \\ \begin{cases}x-5=0\rightarrow x=5 \\ x+2=0{\rightarrow x=-2}\end{cases} \end{gathered}[/tex]
Therefore the solution is {-2, 5}. Correct option: third one.
