d^2=3840

Solution for d^2=3840 equation:

d^2=3840

We move all terms to the left:

d^2-(3840)=0

a = 1; b = 0; c = -3840;
Δ = b2-4ac
Δ = 02-4·1·(-3840)
Δ = 15360
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:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*1}=\frac{0-32\sqrt{15}}{2}
=-\frac{32\sqrt{15}}{2}
=-16\sqrt{15}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*1}=\frac{0+32\sqrt{15}}{2}
=\frac{32\sqrt{15}}{2}
=16\sqrt{15}
$