2/2c-3=3/7c+4

Solution for 2/2c-3=3/7c+4 equation:

2/2c-3=3/7c+4

We move all terms to the left:

2/2c-3-(3/7c+4)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R



Domain of the equation: 7c+4)!=0

c∈R


We get rid of parentheses

2/2c-3/7c-4-3=0

We calculate fractions


14c/14c^2+(-6c)/14c^2-4-3=0

We add all the numbers together, and all the variables

14c/14c^2+(-6c)/14c^2-7=0

We multiply all the terms by the denominator


14c+(-6c)-7*14c^2=0

Wy multiply elements

-98c^2+14c+(-6c)=0

We get rid of parentheses

-98c^2+14c-6c=0

We add all the numbers together, and all the variables

-98c^2+8c=0

a = -98; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-98)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-98}=\frac{-16}{-196}
=4/49
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-98}=\frac{0}{-196}
=0
$