4(1+c)+3-1/3c=1

Solution for 4(1+c)+3-1/3c=1 equation:

4(1+c)+3-1/3c=1

We move all terms to the left:

4(1+c)+3-1/3c-(1)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We add all the numbers together, and all the variables

4(c+1)-1/3c+3-1=0

We add all the numbers together, and all the variables

4(c+1)-1/3c+2=0

We multiply parentheses

4c-1/3c+4+2=0

We multiply all the terms by the denominator


4c*3c+4*3c+2*3c-1=0

Wy multiply elements

12c^2+12c+6c-1=0

We add all the numbers together, and all the variables

12c^2+18c-1=0

a = 12; b = 18; c = -1;
Δ = b2-4ac
Δ = 182-4·12·(-1)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{93}}{2*12}=\frac{-18-2\sqrt{93}}{24}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{93}}{2*12}=\frac{-18+2\sqrt{93}}{24}
$