3c+1/4c=9

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

3c+1/4c=9

We move all terms to the left:

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


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R


We multiply all the terms by the denominator


3c*4c-9*4c+1=0

Wy multiply elements

12c^2-36c+1=0

a = 12; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·12·1
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{78}}{2*12}=\frac{36-4\sqrt{78}}{24}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{78}}{2*12}=\frac{36+4\sqrt{78}}{24}
$