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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-2d+6d-(4d(d-2))-3=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 add all the numbers together, and all the variables

4d-(4d^2-8d)-3=0

We get rid of parentheses

-4d^2+4d+8d-3=0

We add all the numbers together, and all the variables

-4d^2+12d-3=0

a = -4; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·(-4)·(-3)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{6}}{2*-4}=\frac{-12-4\sqrt{6}}{-8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{6}}{2*-4}=\frac{-12+4\sqrt{6}}{-8}
$