(d+3d)-(-d+2)d=3

Solution for (d+3d)-(-d+2)d=3 equation:

(d+3d)-(-d+2)d=3

We move all terms to the left:

(d+3d)-(-d+2)d-(3)=0

We add all the numbers together, and all the variables

(+4d)-(-1d+2)d-3=0

We multiply parentheses

1d^2+(+4d)-2d-3=0

We get rid of parentheses

1d^2+4d-2d-3=0

We add all the numbers together, and all the variables

d^2+2d-3=0

a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*1}=\frac{-6}{2}
=-3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*1}=\frac{2}{2}
=1
$