5v^2+40v+50=0

Solution for 5v^2+40v+50=0 equation:

5v^2+40v+50=0

a = 5; b = 40; c = +50;
Δ = b2-4ac
Δ = 402-4·5·50
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{6}}{2*5}=\frac{-40-10\sqrt{6}}{10}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{6}}{2*5}=\frac{-40+10\sqrt{6}}{10}
$