408=2c+3c2

Solution for 408=2c+3c2 equation:

408=2c+3c^2

We move all terms to the left:

408-(2c+3c^2)=0

We get rid of parentheses

-3c^2-2c+408=0

a = -3; b = -2; c = +408;
Δ = b2-4ac
Δ = -22-4·(-3)·408
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-70}{2*-3}=\frac{-68}{-6}
=11+1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+70}{2*-3}=\frac{72}{-6}
=-12
$