5v^2+20v+10=4v

Solution for 5v^2+20v+10=4v equation:

5v^2+20v+10=4v

We move all terms to the left:

5v^2+20v+10-(4v)=0

We add all the numbers together, and all the variables

5v^2+16v+10=0

a = 5; b = 16; c = +10;
Δ = b2-4ac
Δ = 162-4·5·10
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{14}}{2*5}=\frac{-16-2\sqrt{14}}{10}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{14}}{2*5}=\frac{-16+2\sqrt{14}}{10}
$