3v^2+23v+14=0

Solution for 3v^2+23v+14=0 equation:

3v^2+23v+14=0

a = 3; b = 23; c = +14;
Δ = b2-4ac
Δ = 232-4·3·14
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*3}=\frac{-42}{6}
=-7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*3}=\frac{-4}{6}
=-2/3
$