3/4c+40-1/2c=48

Solution for 3/4c+40-1/2c=48 equation:

3/4c+40-1/2c=48

We move all terms to the left:

3/4c+40-1/2c-(48)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R


We add all the numbers together, and all the variables

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

We calculate fractions


6c/8c^2+(-4c)/8c^2-8=0

We multiply all the terms by the denominator


6c+(-4c)-8*8c^2=0

Wy multiply elements

-64c^2+6c+(-4c)=0

We get rid of parentheses

-64c^2+6c-4c=0

We add all the numbers together, and all the variables

-64c^2+2c=0

a = -64; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-64)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-64}=\frac{-4}{-128}
=1/32
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-64}=\frac{0}{-128}
=0
$