49v^2+42v-16=0

Solution for 49v^2+42v-16=0 equation:

49v^2+42v-16=0

a = 49; b = 42; c = -16;
Δ = b2-4ac
Δ = 422-4·49·(-16)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-70}{2*49}=\frac{-112}{98}
=-1+1/7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+70}{2*49}=\frac{28}{98}
=2/7
$