1/4c=12;c=48

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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} $

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