4=c(3c+2)

Solution for 4=c(3c+2) equation:

4=c(3c+2)

We move all terms to the left:

4-(c(3c+2))=0


We calculate terms in parentheses: -(c(3c+2)), so:

c(3c+2)

We multiply parentheses

3c^2+2c

Back to the equation:

-(3c^2+2c)


We get rid of parentheses

-3c^2-2c+4=0

a = -3; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-3)·4
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*-3}=\frac{2-2\sqrt{13}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*-3}=\frac{2+2\sqrt{13}}{-6}
$