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)+5)=0

We multiply parentheses

12d^2-6d-(4d(d-2)+5)=0


We calculate terms in parentheses: -(4d(d-2)+5), so:

4d(d-2)+5

We multiply parentheses

4d^2-8d+5

Back to the equation:

-(4d^2-8d+5)


We get rid of parentheses

12d^2-4d^2-6d+8d-5=0

We add all the numbers together, and all the variables

8d^2+2d-5=0

a = 8; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·8·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*8}=\frac{-2-2\sqrt{41}}{16}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*8}=\frac{-2+2\sqrt{41}}{16}
$