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

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

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

We move all terms to the left:

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

We multiply parentheses

3c^2+6c-(2(4c+2))=0


We calculate terms in parentheses: -(2(4c+2)), so:

2(4c+2)

We multiply parentheses

8c+4

Back to the equation:

-(8c+4)


We get rid of parentheses

3c^2+6c-8c-4=0

We add all the numbers together, and all the variables

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}
$