# Is there a way to see and understand the steps of reduction while this equation gets solved for V? The result should be somewhere around 6. Thank you.

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Karl on 7 Sep 2022
Commented: Sam Chak on 7 Sep 2022
(65-(50-V))/((15-V*0.5)/(V*0.866)) == ((50-V)-((65/(((15-V*0.5)/(V*0.866))+(1.363)))*(1.363)))/((((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((26.5)-(65-(50-V))/((28.4)/((26.5)-((28.4)/((15-V*0.5)/(V*0.866)))))))+1.732))*1.732+(50-V))-((65/(((15-V*0.5)/(V*0.866))+(1.363)))*(1.363)))/((26.86)-(((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((26.5)-(65-(50-V))/((28.4)/((26.5)-((28.4)/((15-V*0.5)/(V*0.866)))))))+1.732))*1.732+(50-V))-36.6)/((28.4)/((26.86)-((65-(50-V))/1.732)))))

Sam Chak on 7 Sep 2022
Edited: Sam Chak on 7 Sep 2022
The equation is super long with many parentheses and it's hard to interpret without spaces. Maybe you can view it this way:
syms V
eqn = (65-(50-V))/((15-V*0.5)/(V*0.866)) == ((50-V)-((65/(((15-V*0.5)/(V*0.866))+(1.363)))*(1.363)))/((((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((26.5)-(65-(50-V))/((28.4)/((26.5)-((28.4)/((15-V*0.5)/(V*0.866)))))))+1.732))*1.732+(50-V))-((65/(((15-V*0.5)/(V*0.866))+(1.363)))*(1.363)))/((26.86)-(((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((((((50*2-(50-V))/3.05)*1.732+(50-V))-(50-V))/((26.5)-(65-(50-V))/((28.4)/((26.5)-((28.4)/((15-V*0.5)/(V*0.866)))))))+1.732))*1.732+(50-V))-36.6)/((28.4)/((26.86)-((65-(50-V))/1.732)))))
eqn =
sol = solve(eqn)
sol =
The solution requires finding the roots of a 7th-order polynomial equation.
sol = vpasolve(eqn)
sol =
As you can see, one of the five real solutions is , which is probably the one you are referring.
Sam Chak on 7 Sep 2022
@Karl, edited to show the numerical solution of V.