500=2.2v+v^2/20

Solution for 500=2.2v+v^2/20 equation:

500=2.2v+v^2/20

We move all terms to the left:

500-(2.2v+v^2/20)=0

We get rid of parentheses

-v^2/20-2.2v+500=0

We multiply all the terms by the denominator


-v^2-(2.2v)*20+500*20=0

We add all the numbers together, and all the variables

-v^2-(+2.2v)*20+500*20=0

We add all the numbers together, and all the variables

-1v^2-(+2.2v)*20+10000=0

We multiply parentheses

-1v^2-40v+10000=0

a = -1; b = -40; c = +10000;
Δ = b2-4ac
Δ = -402-4·(-1)·10000
Δ = 41600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{41600}=\sqrt{1600*26}=\sqrt{1600}*\sqrt{26}=40\sqrt{26}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40\sqrt{26}}{2*-1}=\frac{40-40\sqrt{26}}{-2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40\sqrt{26}}{2*-1}=\frac{40+40\sqrt{26}}{-2}
$