-0.1d^2+0.5d+0.6=0

Solution for -0.1d^2+0.5d+0.6=0 equation:

-0.1d^2+0.5d+0.6=0

a = -0.1; b = 0.5; c = +0.6;
Δ = b2-4ac
Δ = 0.52-4·(-0.1)·0.6
Δ = 0.49
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{-(0.5)-\sqrt{0.49}}{2*-0.1}=\frac{-0.5-\sqrt{0.49}}{-0.2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.49}}{2*-0.1}=\frac{-0.5+\sqrt{0.49}}{-0.2}
$