(4/d-5)-(9/4d+8)+(2/d-5)=0

Solution for (4/d-5)-(9/4d+8)+(2/d-5)=0 equation:

(4/d-5)-(9/4d+8)+(2/d-5)=0


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

d∈R



Domain of the equation: 4d+8)!=0

d∈R


We get rid of parentheses

4/d-9/4d+2/d-5-8-5=0

We calculate fractions


(8d+4)/4d^2+(-9d)/4d^2-5-8-5=0

We add all the numbers together, and all the variables

(8d+4)/4d^2+(-9d)/4d^2-18=0

We multiply all the terms by the denominator


(8d+4)+(-9d)-18*4d^2=0

Wy multiply elements

-72d^2+(8d+4)+(-9d)=0

We get rid of parentheses

-72d^2+8d-9d+4=0

We add all the numbers together, and all the variables

-72d^2-1d+4=0

a = -72; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·(-72)·4
Δ = 1153
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{-(-1)-\sqrt{1153}}{2*-72}=\frac{1-\sqrt{1153}}{-144}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1153}}{2*-72}=\frac{1+\sqrt{1153}}{-144}
$