2d+2d2=4d+2

Solution for 2d+2d2=4d+2 equation:

2d+2d^2=4d+2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2d^2-2d-2=0

a = 2; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·2·(-2)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*2}=\frac{2-2\sqrt{5}}{4}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*2}=\frac{2+2\sqrt{5}}{4}
$