3d+(5-d);d=4

Solution for 3d+(5-d);d=4 equation:

3d+(5-d)d=4

We move all terms to the left:

3d+(5-d)d-(4)=0

We add all the numbers together, and all the variables

3d+(-1d+5)d-4=0

We multiply parentheses

-1d^2+3d+5d-4=0

We add all the numbers together, and all the variables

-1d^2+8d-4=0

a = -1; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·(-1)·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*-1}=\frac{-8-4\sqrt{3}}{-2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*-1}=\frac{-8+4\sqrt{3}}{-2}
$