I think we have a confusion about the meaning of solve block.
A solve block is that construct that starts with the statement "Given" followed by equations and constraints and usually end with "Find(..)" (while also minerr, maximize, odesolve,.. are possible).
You can assign the output of Find() to a variable. Typically such a solve block is evaluated numerically - you type something like "Solution:=Find(x)" and Mathcad would use a numerical iteration algorithm to get close to a solution. It needs a guess value for the searched for variable to do so and this is what you didn't like in a prior post.
But you can also evaluate a solveblock symbolically by typing "Solution:=Find(x)-->" Now Mathcad doesn't need a guess and will use "exact" symbolical ways to find the solutions. Unfortunately usually if the symbolical solve command fails a symbolical evaluated solve block will fail too.
I think you thought we mean the symbolical solve command you used in conjunction with assume, etc, when we were talking about a solve block.
Some questions about your problem:
You say that you would just use the magnitude of the solution of the quadratic equation.
1) What if the magnitudes of both solutions are inbetween v0 and v1?
2) If a solution is neagtive we would have to use its absolute value in comparing with v0 and v1 as ist only the magnitude which counts, right?
3) In the first sheet you posted both v0 and v1 were negative! How would you expect the magnitude of any number being inbetween??? Are you sure you will use the magnitude (which is positive or zero) or would you rather use the real part of the solution (which can be negative, too)??
