2/3c=8;c=9

Solution for 2/3c=8;c=9 equation:

2/3c=8c=9

We move all terms to the left:

2/3c-(8c)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We add all the numbers together, and all the variables

-8c+2/3c=0

We multiply all the terms by the denominator


-8c*3c+2=0

Wy multiply elements

-24c^2+2=0

a = -24; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-24)·2
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*-24}=\frac{0-8\sqrt{3}}{-48}
=-\frac{8\sqrt{3}}{-48}
=-\frac{\sqrt{3}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-24}=\frac{0+8\sqrt{3}}{-48}
=\frac{8\sqrt{3}}{-48}
=\frac{\sqrt{3}}{-6}
$