-4v^2+12v+112=0

Solution for -4v^2+12v+112=0 equation:

-4v^2+12v+112=0

a = -4; b = 12; c = +112;
Δ = b2-4ac
Δ = 122-4·(-4)·112
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{-(12)-44}{2*-4}=\frac{-56}{-8}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+44}{2*-4}=\frac{32}{-8}
=-4
$