d.0.2(d-6)=0.3d+5-3+0.1d

Solution for d.0.2(d-6)=0.3d+5-3+0.1d equation:

d.0.2(d-6)=0.3d+5-3+0.1d

We move all terms to the left:

d.0.2(d-6)-(0.3d+5-3+0.1d)=0

We add all the numbers together, and all the variables

d.0.2(d-6)-(0.4d+2)=0

We multiply parentheses

d^2-6d-(0.4d+2)=0

We get rid of parentheses

d^2-6d-0.4d-2=0

We add all the numbers together, and all the variables

d^2-6.4d-2=0

a = 1; b = -6.4; c = -2;
Δ = b2-4ac
Δ = -6.42-4·1·(-2)
Δ = 48.96
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{-(-6.4)-\sqrt{48.96}}{2*1}=\frac{6.4-\sqrt{48.96}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6.4)+\sqrt{48.96}}{2*1}=\frac{6.4+\sqrt{48.96}}{2}
$