3(2d-1)2d=4d(d-2)=5

Solution for 3(2d-1)2d=4d(d-2)=5 equation:

3(2d-1)2d=4d(d-2)=5

We move all terms to the left:

3(2d-1)2d-(4d(d-2))=0

We multiply parentheses

12d^2-6d-(4d(d-2))=0


We calculate terms in parentheses: -(4d(d-2)), so:

4d(d-2)

We multiply parentheses

4d^2-8d

Back to the equation:

-(4d^2-8d)


We get rid of parentheses

12d^2-4d^2-6d+8d=0

We add all the numbers together, and all the variables

8d^2+2d=0

a = 8; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·8·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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