3(c+2)c+1=57

Solution for 3(c+2)c+1=57 equation:

3(c+2)c+1=57

We move all terms to the left:

3(c+2)c+1-(57)=0

We add all the numbers together, and all the variables

3(c+2)c-56=0

We multiply parentheses

3c^2+6c-56=0

a = 3; b = 6; c = -56;
Δ = b2-4ac
Δ = 62-4·3·(-56)
Δ = 708
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:

$\sqrt{\Delta}=\sqrt{708}=\sqrt{4*177}=\sqrt{4}*\sqrt{177}=2\sqrt{177}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{177}}{2*3}=\frac{-6-2\sqrt{177}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{177}}{2*3}=\frac{-6+2\sqrt{177}}{6}
$