4d^2+56d+96=0

Solution for 4d^2+56d+96=0 equation:

4d^2+56d+96=0

a = 4; b = 56; c = +96;
Δ = b2-4ac
Δ = 562-4·4·96
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-40}{2*4}=\frac{-96}{8}
=-12
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+40}{2*4}=\frac{-16}{8}
=-2
$