3v^2+20v+25=0

Solution for 3v^2+20v+25=0 equation:

3v^2+20v+25=0

a = 3; b = 20; c = +25;
Δ = b2-4ac
Δ = 202-4·3·25
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*3}=\frac{-30}{6}
=-5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*3}=\frac{-10}{6}
=-1+2/3
$