4d+(2d-2)d=6

Solution for 4d+(2d-2)d=6 equation:

4d+(2d-2)d=6

We move all terms to the left:

4d+(2d-2)d-(6)=0

We multiply parentheses

2d^2+4d-2d-6=0

We add all the numbers together, and all the variables

2d^2+2d-6=0

a = 2; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·2·(-6)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*2}=\frac{-2-2\sqrt{13}}{4}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*2}=\frac{-2+2\sqrt{13}}{4}
$