1/2c+7=9+3/4c

Solution for 1/2c+7=9+3/4c equation:

1/2c+7=9+3/4c

We move all terms to the left:

1/2c+7-(9+3/4c)=0


Domain of the equation: 2c!=0

c!=0/2

c!=0

c∈R



Domain of the equation: 4c)!=0

c!=0/1

c!=0

c∈R


We add all the numbers together, and all the variables

1/2c-(3/4c+9)+7=0

We get rid of parentheses

1/2c-3/4c-9+7=0

We calculate fractions


4c/8c^2+(-6c)/8c^2-9+7=0

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Wy multiply elements

-16c^2+4c+(-6c)=0

We get rid of parentheses

-16c^2+4c-6c=0

We add all the numbers together, and all the variables

-16c^2-2c=0

a = -16; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-16)·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*-16}=\frac{0}{-32}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-16}=\frac{4}{-32}
=-1/8
$