4/2d-5=3d-2/2d-5-2

Solution for 4/2d-5=3d-2/2d-5-2 equation:

4/2d-5=3d-2/2d-5-2

We move all terms to the left:

4/2d-5-(3d-2/2d-5-2)=0


Domain of the equation: 2d!=0

d!=0/2

d!=0

d∈R



Domain of the equation: 2d-5-2)!=0

We move all terms containing d to the left, all other terms to the right

2d-2)!=5

d∈R


We add all the numbers together, and all the variables

4/2d-(3d-2/2d-7)-5=0

We get rid of parentheses

4/2d-3d+2/2d+7-5=0

We multiply all the terms by the denominator


-3d*2d+7*2d-5*2d+4+2=0

We add all the numbers together, and all the variables

-3d*2d+7*2d-5*2d+6=0

Wy multiply elements

-6d^2+14d-10d+6=0

We add all the numbers together, and all the variables

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

a = -6; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-6)·6
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*-6}=\frac{-4-4\sqrt{10}}{-12}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*-6}=\frac{-4+4\sqrt{10}}{-12}
$