1/2c+13=13c+48

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

1/2c+13=13c+48

We move all terms to the left:

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


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-26c^2-70c+1=0

a = -26; b = -70; c = +1;
Δ = b2-4ac
Δ = -702-4·(-26)·1
Δ = 5004
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:

$\sqrt{\Delta}=\sqrt{5004}=\sqrt{36*139}=\sqrt{36}*\sqrt{139}=6\sqrt{139}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-6\sqrt{139}}{2*-26}=\frac{70-6\sqrt{139}}{-52}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+6\sqrt{139}}{2*-26}=\frac{70+6\sqrt{139}}{-52}
$