0=-0,004v^2+4/5v-20,4

Solution for 0=-0,004v^2+4/5v-20,4 equation:

0=-0.004v^2+4/5v-20.4

We move all terms to the left:

0-(-0.004v^2+4/5v-20.4)=0


Domain of the equation: 5v-20.4)!=0

v∈R


We add all the numbers together, and all the variables

-(-0.004v^2+4/5v-20.4)=0

We get rid of parentheses

0.004v^2-4/5v+20.4=0

We multiply all the terms by the denominator


(0.004v^2)*5v+(20.4)*5v-4=0

We multiply parentheses

0v^2+102v-4=0

We add all the numbers together, and all the variables

v^2+102v-4=0

a = 1; b = 102; c = -4;
Δ = b2-4ac
Δ = 1022-4·1·(-4)
Δ = 10420
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:

$\sqrt{\Delta}=\sqrt{10420}=\sqrt{4*2605}=\sqrt{4}*\sqrt{2605}=2\sqrt{2605}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-2\sqrt{2605}}{2*1}=\frac{-102-2\sqrt{2605}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+2\sqrt{2605}}{2*1}=\frac{-102+2\sqrt{2605}}{2}
$