(-0.0032d^2)+d+3=0

Solution for (-0.0032d^2)+d+3=0 equation:

(-0.0032d^2)+d+3=0

We get rid of parentheses

-0.0032d^2+d+3=0

a = -0.0032; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-0.0032)·3
Δ = 1.0384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1.0384}}{2*-0.0032}=\frac{-1-\sqrt{1.0384}}{-0.0064}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.0384}}{2*-0.0032}=\frac{-1+\sqrt{1.0384}}{-0.0064}
$