Circle describe and correct each error 10v^2-46v=2010v^2-46v-20=02(5v^2-23v-10)=02(v-25)(c+2)=0V-25=0 and v+2=0V=25 and v=-2

Given:
The solved quadratic equation is given.
Required:
To describe and correct each error.
Answer:
Let us find the the error.
[tex]\begin{gathered} 10\nu^2-46\nu=20 \\ 10\nu^2-46\nu-20=0 \\ 2(5\nu^2-23\nu-10)=0 \\ 5\nu^2-23\nu-10=0 \\ 5\nu^2-25\nu+2\nu-10=0 \\ 5\nu(\nu-5)+2(\nu-5)=0 \\ (\nu-5)(5\nu+2)=0 \\ (\nu-5)=0,(5\nu+2)=0 \\ \nu=5,5\nu=-2 \\ \nu=5,\nu=\frac{-2}{5} \end{gathered}[/tex]Hence, the error is
[tex](\nu-25),(\nu+2)[/tex]There should be
[tex](\nu-5),(5\nu+2)[/tex]in the calculation.
Final Answer:
The error is
[tex](\nu-25),(\nu+2)[/tex]