1/3c+2=7,c=

Solution for 1/3c+2=7,c= equation:

1/3c+2=7.c=

We move all terms to the left:

1/3c+2-(7.c)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We add all the numbers together, and all the variables

1/3c-(+7.c)+2=0

We get rid of parentheses

1/3c-7.c+2=0

We multiply all the terms by the denominator


-(7.c)*3c+2*3c+1=0

We add all the numbers together, and all the variables

-(+7.c)*3c+2*3c+1=0

We multiply parentheses

-21c^2+2*3c+1=0

Wy multiply elements

-21c^2+6c+1=0

a = -21; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-21)·1
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{30}}{2*-21}=\frac{-6-2\sqrt{30}}{-42}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{30}}{2*-21}=\frac{-6+2\sqrt{30}}{-42}
$