3v^2+28v+39=0

Solution for 3v^2+28v+39=0 equation:

3v^2+28v+39=0

a = 3; b = 28; c = +39;
Δ = b2-4ac
Δ = 282-4·3·39
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{79}}{2*3}=\frac{-28-2\sqrt{79}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{79}}{2*3}=\frac{-28+2\sqrt{79}}{6}
$