1/4c=12;c=48

Solution for 1/4c=12;c=48 equation:

1/4c=12c=48

We move all terms to the left:

1/4c-(12c)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R


We add all the numbers together, and all the variables

-12c+1/4c=0

We multiply all the terms by the denominator


-12c*4c+1=0

Wy multiply elements

-48c^2+1=0

a = -48; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-48)·1
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-48}=\frac{0-8\sqrt{3}}{-96}
=-\frac{8\sqrt{3}}{-96}
=-\frac{\sqrt{3}}{-12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-48}=\frac{0+8\sqrt{3}}{-96}
=\frac{8\sqrt{3}}{-96}
=\frac{\sqrt{3}}{-12}
$