7/2c=5+2c+3

Solution for 7/2c=5+2c+3 equation:

7/2c=5+2c+3

We move all terms to the left:

7/2c-(5+2c+3)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We add all the numbers together, and all the variables

7/2c-(2c+8)=0

We get rid of parentheses

7/2c-2c-8=0

We multiply all the terms by the denominator


-2c*2c-8*2c+7=0

Wy multiply elements

-4c^2-16c+7=0

a = -4; b = -16; c = +7;
Δ = b2-4ac
Δ = -162-4·(-4)·7
Δ = 368
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{23}}{2*-4}=\frac{16-4\sqrt{23}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{23}}{2*-4}=\frac{16+4\sqrt{23}}{-8}
$