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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

0.2d(d-6)+0.3d-(0.1d+2)=0

We add all the numbers together, and all the variables

0.3d+0.2d(d-6)-(0.1d+2)=0

We multiply parentheses

0d^2+0.3d+0d-(0.1d+2)=0

We get rid of parentheses

0d^2+0.3d+0d-0.1d-2=0

We add all the numbers together, and all the variables

d^2+1.2d-2=0

a = 1; b = 1.2; c = -2;
Δ = b2-4ac
Δ = 1.22-4·1·(-2)
Δ = 9.44
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{-(1.2)-\sqrt{9.44}}{2*1}=\frac{-1.2-\sqrt{9.44}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{9.44}}{2*1}=\frac{-1.2+\sqrt{9.44}}{2}
$