1/2c+13=3c+48

Solution for 1/2c+13=3c+48 equation:

1/2c+13=3c+48

We move all terms to the left:

1/2c+13-(3c+48)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We get rid of parentheses

1/2c-3c-48+13=0

We multiply all the terms by the denominator


-3c*2c-48*2c+13*2c+1=0

Wy multiply elements

-6c^2-96c+26c+1=0

We add all the numbers together, and all the variables

-6c^2-70c+1=0

a = -6; b = -70; c = +1;
Δ = b2-4ac
Δ = -702-4·(-6)·1
Δ = 4924
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:

$\sqrt{\Delta}=\sqrt{4924}=\sqrt{4*1231}=\sqrt{4}*\sqrt{1231}=2\sqrt{1231}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-2\sqrt{1231}}{2*-6}=\frac{70-2\sqrt{1231}}{-12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{1231}}{2*-6}=\frac{70+2\sqrt{1231}}{-12}
$