10=-0.2d^2+3d+6

Solution for 10=-0.2d^2+3d+6 equation:

10=-0.2d^2+3d+6

We move all terms to the left:

10-(-0.2d^2+3d+6)=0

We get rid of parentheses

0.2d^2-3d-6+10=0

We add all the numbers together, and all the variables

0.2d^2-3d+4=0

a = 0.2; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·0.2·4
Δ = 5.8
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{-(-3)-\sqrt{5.8}}{2*0.2}=\frac{3-\sqrt{5.8}}{0.4}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5.8}}{2*0.2}=\frac{3+\sqrt{5.8}}{0.4}
$