(d+5d)-(-6d+2);d=3

Solution for (d+5d)-(-6d+2);d=3 equation:

(d+5d)-(-6d+2)d=3

We move all terms to the left:

(d+5d)-(-6d+2)d-(3)=0

We add all the numbers together, and all the variables

(+6d)-(-6d+2)d-3=0

We multiply parentheses

6d^2+(+6d)-2d-3=0

We get rid of parentheses

6d^2+6d-2d-3=0

We add all the numbers together, and all the variables

6d^2+4d-3=0

a = 6; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·6·(-3)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{22}}{2*6}=\frac{-4-2\sqrt{22}}{12}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{22}}{2*6}=\frac{-4+2\sqrt{22}}{12}
$