3(2d^2)-4d=(2d)^2

Solution for 3(2d^2)-4d=(2d)^2 equation:

3(2d^2)-4d=(2d)^2

We move all terms to the left:

3(2d^2)-4d-((2d)^2)=0

determiningTheFunctionDomain
32d^2-4d-2d^2=0

We add all the numbers together, and all the variables

32d^2-2d^2-4d=0

We add all the numbers together, and all the variables

30d^2-4d=0

a = 30; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·30·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*30}=\frac{0}{60}
=0
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*30}=\frac{8}{60}
=2/15
$