1/2c-19=c+1

Solution for 1/2c-19=c+1 equation:

1/2c-19=c+1

We move all terms to the left:

1/2c-19-(c+1)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We get rid of parentheses

1/2c-c-1-19=0

We multiply all the terms by the denominator


-c*2c-1*2c-19*2c+1=0

Wy multiply elements

-2c^2-2c-38c+1=0

We add all the numbers together, and all the variables

-2c^2-40c+1=0

a = -2; b = -40; c = +1;
Δ = b2-4ac
Δ = -402-4·(-2)·1
Δ = 1608
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:

$\sqrt{\Delta}=\sqrt{1608}=\sqrt{4*402}=\sqrt{4}*\sqrt{402}=2\sqrt{402}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{402}}{2*-2}=\frac{40-2\sqrt{402}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{402}}{2*-2}=\frac{40+2\sqrt{402}}{-4}
$