d^2+(3d)^2=20^2

Solution for d^2+(3d)^2=20^2 equation:

d^2+(3d)^2=20^2

We move all terms to the left:

d^2+(3d)^2-(20^2)=0

We add all the numbers together, and all the variables

4d^2-400=0

a = 4; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·4·(-400)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*4}=\frac{-80}{8}
=-10
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*4}=\frac{80}{8}
=10
$