9+3/4c=c-9

Solution for 9+3/4c=c-9 equation:

9+3/4c=c-9

We move all terms to the left:

9+3/4c-(c-9)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R


We get rid of parentheses

3/4c-c+9+9=0

We multiply all the terms by the denominator


-c*4c+9*4c+9*4c+3=0

Wy multiply elements

-4c^2+36c+36c+3=0

We add all the numbers together, and all the variables

-4c^2+72c+3=0

a = -4; b = 72; c = +3;
Δ = b2-4ac
Δ = 722-4·(-4)·3
Δ = 5232
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:

$\sqrt{\Delta}=\sqrt{5232}=\sqrt{16*327}=\sqrt{16}*\sqrt{327}=4\sqrt{327}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-4\sqrt{327}}{2*-4}=\frac{-72-4\sqrt{327}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+4\sqrt{327}}{2*-4}=\frac{-72+4\sqrt{327}}{-8}
$