C(t)=-3t^2+12t

Solution for C(t)=-3t^2+12t equation:

(C)=-3C^2+12C

We move all terms to the left:

(C)-(-3C^2+12C)=0

We get rid of parentheses

3C^2-12C+C=0

We add all the numbers together, and all the variables

3C^2-11C=0

a = 3; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·3·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$C_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*3}=\frac{0}{6}
=0
$
$C_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*3}=\frac{22}{6}
=3+2/3
$