3v^2+27v+49=0

Solution for 3v^2+27v+49=0 equation:

3v^2+27v+49=0

a = 3; b = 27; c = +49;
Δ = b2-4ac
Δ = 272-4·3·49
Δ = 141
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-\sqrt{141}}{2*3}=\frac{-27-\sqrt{141}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{141}}{2*3}=\frac{-27+\sqrt{141}}{6}
$