-3/4c-8=1/8c+20

Solution for -3/4c-8=1/8c+20 equation:

-3/4c-8=1/8c+20

We move all terms to the left:

-3/4c-8-(1/8c+20)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



Domain of the equation: 8c+20)!=0

c∈R


We get rid of parentheses

-3/4c-1/8c-20-8=0

We calculate fractions


(-24c)/32c^2+(-4c)/32c^2-20-8=0

We add all the numbers together, and all the variables

(-24c)/32c^2+(-4c)/32c^2-28=0

We multiply all the terms by the denominator


(-24c)+(-4c)-28*32c^2=0

Wy multiply elements

-896c^2+(-24c)+(-4c)=0

We get rid of parentheses

-896c^2-24c-4c=0

We add all the numbers together, and all the variables

-896c^2-28c=0

a = -896; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·(-896)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*-896}=\frac{0}{-1792}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*-896}=\frac{56}{-1792}
=-1/32
$