d^2=12d+28

Solution for d^2=12d+28 equation:

d^2=12d+28

We move all terms to the left:

d^2-(12d+28)=0

We get rid of parentheses

d^2-12d-28=0

a = 1; b = -12; c = -28;
Δ = b2-4ac
Δ = -122-4·1·(-28)
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{-(-12)-16}{2*1}=\frac{-4}{2}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*1}=\frac{28}{2}
=14
$