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

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

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

We move all terms to the left:

3(2d-1)-(4d(d-2)+5)=0

We multiply parentheses

6d-(4d(d-2)+5)-3=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

-4d^2+6d+8d-5-3=0

We add all the numbers together, and all the variables

-4d^2+14d-8=0

a = -4; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·(-4)·(-8)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{17}}{2*-4}=\frac{-14-2\sqrt{17}}{-8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{17}}{2*-4}=\frac{-14+2\sqrt{17}}{-8}
$