3v^2+25v+50=0

Solution for 3v^2+25v+50=0 equation:

3v^2+25v+50=0

a = 3; b = 25; c = +50;
Δ = b2-4ac
Δ = 252-4·3·50
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5}{2*3}=\frac{-30}{6}
=-5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5}{2*3}=\frac{-20}{6}
=-3+1/3
$