5v2+-41v+42=0

Solution for 5v2+-41v+42=0 equation:

5v^2+-41v+42=0

We add all the numbers together, and all the variables

5v^2-41v=0

a = 5; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·5·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*5}=\frac{0}{10}
=0
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*5}=\frac{82}{10}
=8+1/5
$