t=-3t2+12t

Solution for t=-3t2+12t equation:

t=-3t^2+12t

We move all terms to the left:

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

We get rid of parentheses

3t^2-12t+t=0

We add all the numbers together, and all the variables

3t^2-11t=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:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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