3/2c-3c+4=5/2c+7+3

Solution for 3/2c-3c+4=5/2c+7+3 equation:

3/2c-3c+4=5/2c+7+3

We move all terms to the left:

3/2c-3c+4-(5/2c+7+3)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R



Domain of the equation: 2c+7+3)!=0

We move all terms containing c to the left, all other terms to the right

2c+3)!=-7

c∈R


We add all the numbers together, and all the variables

3/2c-3c-(5/2c+10)+4=0

We add all the numbers together, and all the variables

-3c+3/2c-(5/2c+10)+4=0

We get rid of parentheses

-3c+3/2c-5/2c-10+4=0

We multiply all the terms by the denominator


-3c*2c-10*2c+4*2c+3-5=0

We add all the numbers together, and all the variables

-3c*2c-10*2c+4*2c-2=0

Wy multiply elements

-6c^2-20c+8c-2=0

We add all the numbers together, and all the variables

-6c^2-12c-2=0

a = -6; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·(-6)·(-2)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*-6}=\frac{12-4\sqrt{6}}{-12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*-6}=\frac{12+4\sqrt{6}}{-12}
$