35v^2+54v+7=0

Solution for 35v^2+54v+7=0 equation:

35v^2+54v+7=0

a = 35; b = 54; c = +7;
Δ = b2-4ac
Δ = 542-4·35·7
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-44}{2*35}=\frac{-98}{70}
=-1+2/5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+44}{2*35}=\frac{-10}{70}
=-1/7
$