3v^2+42v+72=0

Solution for 3v^2+42v+72=0 equation:

3v^2+42v+72=0

a = 3; b = 42; c = +72;
Δ = b2-4ac
Δ = 422-4·3·72
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-30}{2*3}=\frac{-72}{6}
=-12
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+30}{2*3}=\frac{-12}{6}
=-2
$