2c(11/3)=6

Solution for 2c(11/3)=6 equation:

2c(11/3)=6

We move all terms to the left:

2c(11/3)-(6)=0

We add all the numbers together, and all the variables

2c(+11/3)-6=0

We multiply parentheses

22c^2-6=0

a = 22; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·22·(-6)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{33}}{2*22}=\frac{0-4\sqrt{33}}{44}
=-\frac{4\sqrt{33}}{44}
=-\frac{\sqrt{33}}{11}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{33}}{2*22}=\frac{0+4\sqrt{33}}{44}
=\frac{4\sqrt{33}}{44}
=\frac{\sqrt{33}}{11}
$