4/3c-8=9/c

Solution for 4/3c-8=9/c equation:

4/3c-8=9/c

We move all terms to the left:

4/3c-8-(9/c)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R



Domain of the equation: c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

4/3c-(+9/c)-8=0

We get rid of parentheses

4/3c-9/c-8=0

We calculate fractions


4c/3c^2+(-27c)/3c^2-8=0

We multiply all the terms by the denominator


4c+(-27c)-8*3c^2=0

Wy multiply elements

-24c^2+4c+(-27c)=0

We get rid of parentheses

-24c^2+4c-27c=0

We add all the numbers together, and all the variables

-24c^2-23c=0

a = -24; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·(-24)·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*-24}=\frac{0}{-48}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*-24}=\frac{46}{-48}
=-23/24
$