2/d-5-9/4d+9+3/d-5=0

Solution for 2/d-5-9/4d+9+3/d-5=0 equation:

2/d-5-9/4d+9+3/d-5=0


Domain of the equation: d!=0

d∈R



Domain of the equation: 4d!=0

d!=0/4

d!=0

d∈R


We add all the numbers together, and all the variables

2/d-9/4d+3/d-1=0

We calculate fractions


(12d+2)/4d^2+(-9d)/4d^2-1=0

We multiply all the terms by the denominator


(12d+2)+(-9d)-1*4d^2=0

Wy multiply elements

-4d^2+(12d+2)+(-9d)=0

We get rid of parentheses

-4d^2+12d-9d+2=0

We add all the numbers together, and all the variables

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

a = -4; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-4)·2
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*-4}=\frac{-3-\sqrt{41}}{-8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*-4}=\frac{-3+\sqrt{41}}{-8}
$