7(d-3)+3d(d-3)=12

Solution for 7(d-3)+3d(d-3)=12 equation:

7(d-3)+3d(d-3)=12

We move all terms to the left:

7(d-3)+3d(d-3)-(12)=0

We multiply parentheses

3d^2+7d-9d-21-12=0

We add all the numbers together, and all the variables

3d^2-2d-33=0

a = 3; b = -2; c = -33;
Δ = b2-4ac
Δ = -22-4·3·(-33)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-20}{2*3}=\frac{-18}{6}
=-3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+20}{2*3}=\frac{22}{6}
=3+2/3
$