16+d*(3d+d)=32

Solution for 16+d*(3d+d)=32 equation:

16+d(3d+d)=32

We move all terms to the left:

16+d(3d+d)-(32)=0

We add all the numbers together, and all the variables

d(+4d)+16-32=0

We add all the numbers together, and all the variables

d(+4d)-16=0

We multiply parentheses

4d^2-16=0

a = 4; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·4·(-16)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*4}=\frac{-16}{8}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*4}=\frac{16}{8}
=2
$