Factoring out an equation I found this equation in a book:
$$ m_0 v_0^2 + m_1 v_1^2 = m_0 v_{0_{Final}}^2 + m_1 v_{1_{Final}}^2.$$
It says that

Notice that you have a different equation with the same two unknown variables: $v_{0_{Final}}$ and $v_{1_{Final}}$.
  You can now factor these out and come up with a single equation for each unknown. These are the
  formulas that you end up with when all is done

$$v_{0_{Final}} = \frac{(m_0 - m_1) v_0 + 2 m_1 v_1 }{m_0 + m_1} \\    
v_{1_{Final}} = \frac{(m_1 - m_0) v_1 + 2 m_0 v_0 }{ m_0 + m_1}.
$$
Can someone please break this result down step-by-step to solve for these 2 unknown variables?
 A: I believe you need another equation to solve logically. The other equation is probably:
$$m_0v_0+m_1v_1=m_0v_{0_{Final}}+m_1v_{1_{Final}}$$
(conservation of momentum).
Without this equation no one can say what $v_{0_{Final}}$ and $v_{1_{Final}}$ should be, as there are infinitely many solutions for example including $v_{0_{Final}}=v_0$ and $v_{1_{Final}}=v_1$ as you already thought.
A: There is another easier way of solving it than the quadratic, (easier at least according to me). This is the solution:
(I am using $u_1$ and $u_2$ for the initial velocities and $v_1$ and $v_2$ for the final velocities.)
$$m_1u_1+m_2u_2=m_1v_1+m_2v_2\tag{Cons. of momentum}$$
$$m_1(u_1-v_1)=m_2(v_2-u_2)\tag{1}$$
$$m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2\tag{Cons. of Energy}$$
$$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$$
$$m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(u_2+v_2)$$
Dividing by (1): 
$$u_1+v_1=u_2+v_2 \tag{A}$$
[This equation can also be stated as: $u_1-u_2=v_2-v_1$, i.e. Velocity of approach = Velocity of seperation.]
And from (1):
$$m_1u_1-m_1v_1=m_2v_2-m_2u_2 \tag{B}$$
(A) and (B) are two linear equations in two unknowns, $v_1$ and $v_2$. Solving by elimination, we get:
$$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_1}{m_1+m_2}u_2$$$$v_2=\frac{2m_2}{m1+m2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2$$
