3v^2+28v=49=0

Solution for 3v^2+28v=49=0 equation:

3v^2+28v=49=0

We move all terms to the left:

3v^2+28v-(49)=0

a = 3; b = 28; c = -49;
Δ = b2-4ac
Δ = 282-4·3·(-49)
Δ = 1372
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:

$\sqrt{\Delta}=\sqrt{1372}=\sqrt{196*7}=\sqrt{196}*\sqrt{7}=14\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-14\sqrt{7}}{2*3}=\frac{-28-14\sqrt{7}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+14\sqrt{7}}{2*3}=\frac{-28+14\sqrt{7}}{6}
$