1/5c+2=1/7c+6

Solution for 1/5c+2=1/7c+6 equation:

1/5c+2=1/7c+6

We move all terms to the left:

1/5c+2-(1/7c+6)=0


Domain of the equation: 5c!=0

c!=0/5

c!=0

c∈R



Domain of the equation: 7c+6)!=0

c∈R


We get rid of parentheses

1/5c-1/7c-6+2=0

We calculate fractions


7c/35c^2+(-5c)/35c^2-6+2=0

We add all the numbers together, and all the variables

7c/35c^2+(-5c)/35c^2-4=0

We multiply all the terms by the denominator


7c+(-5c)-4*35c^2=0

Wy multiply elements

-140c^2+7c+(-5c)=0

We get rid of parentheses

-140c^2+7c-5c=0

We add all the numbers together, and all the variables

-140c^2+2c=0

a = -140; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-140)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-140}=\frac{-4}{-280}
=1/70
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-140}=\frac{0}{-280}
=0
$