-0.1d^2+d+.5=0

Solution for -0.1d^2+d+.5=0 equation:

-0.1d^2+d+.5=0

We add all the numbers together, and all the variables

-0.1d^2+d+0.5=0

a = -0.1; b = 1; c = +0.5;
Δ = b2-4ac
Δ = 12-4·(-0.1)·0.5
Δ = 1.2
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{1.2}}{2*-0.1}=\frac{-1-\sqrt{1.2}}{-0.2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.2}}{2*-0.1}=\frac{-1+\sqrt{1.2}}{-0.2}
$