2v^2+3v-104=0

Solution for 2v^2+3v-104=0 equation:

2v^2+3v-104=0

a = 2; b = 3; c = -104;
Δ = b2-4ac
Δ = 32-4·2·(-104)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-29}{2*2}=\frac{-32}{4}
=-8
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+29}{2*2}=\frac{26}{4}
=6+1/2
$