-3/4c+1/3c=8/2

Solution for -3/4c+1/3c=8/2 equation:

-3/4c+1/3c=8/2

We move all terms to the left:

-3/4c+1/3c-(8/2)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We add all the numbers together, and all the variables

-3/4c+1/3c-4=0

We calculate fractions


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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

-48c^2+4c+(-9c)=0

We get rid of parentheses

-48c^2+4c-9c=0

We add all the numbers together, and all the variables

-48c^2-5c=0

a = -48; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-48)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-48}=\frac{0}{-96}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-48}=\frac{10}{-96}
=-5/48
$