14v^2+96v-128=0

Solution for 14v^2+96v-128=0 equation:

14v^2+96v-128=0

a = 14; b = 96; c = -128;
Δ = b2-4ac
Δ = 962-4·14·(-128)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-128}{2*14}=\frac{-224}{28}
=-8
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+128}{2*14}=\frac{32}{28}
=1+1/7
$