1/2c-8=3+4c

Solution for 1/2c-8=3+4c equation:

1/2c-8=3+4c

We move all terms to the left:

1/2c-8-(3+4c)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We add all the numbers together, and all the variables

1/2c-(4c+3)-8=0

We get rid of parentheses

1/2c-4c-3-8=0

We multiply all the terms by the denominator


-4c*2c-3*2c-8*2c+1=0

Wy multiply elements

-8c^2-6c-16c+1=0

We add all the numbers together, and all the variables

-8c^2-22c+1=0

a = -8; b = -22; c = +1;
Δ = b2-4ac
Δ = -222-4·(-8)·1
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{129}}{2*-8}=\frac{22-2\sqrt{129}}{-16}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{129}}{2*-8}=\frac{22+2\sqrt{129}}{-16}
$