3c-45c=543c/4c+54

Solution for 3c-45c=543c/4c+54 equation:

3c-45c=543c/4c+54

We move all terms to the left:

3c-45c-(543c/4c+54)=0


Domain of the equation: 4c+54)!=0

c∈R


We add all the numbers together, and all the variables

-42c-(543c/4c+54)=0

We get rid of parentheses

-42c-543c/4c-54=0

We multiply all the terms by the denominator


-42c*4c-543c-54*4c=0

We add all the numbers together, and all the variables

-543c-42c*4c-54*4c=0

Wy multiply elements

-168c^2-543c-216c=0

We add all the numbers together, and all the variables

-168c^2-759c=0

a = -168; b = -759; c = 0;
Δ = b2-4ac
Δ = -7592-4·(-168)·0
Δ = 576081
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576081}=759$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-759)-759}{2*-168}=\frac{0}{-336}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-759)+759}{2*-168}=\frac{1518}{-336}
=-4+29/56
$