t=-3t^2+24t+6

Solution for t=-3t^2+24t+6 equation:

t=-3t^2+24t+6

We move all terms to the left:

t-(-3t^2+24t+6)=0

We get rid of parentheses

3t^2-24t+t-6=0

We add all the numbers together, and all the variables

3t^2-23t-6=0

a = 3; b = -23; c = -6;
Δ = b2-4ac
Δ = -232-4·3·(-6)
Δ = 601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{601}}{2*3}=\frac{23-\sqrt{601}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{601}}{2*3}=\frac{23+\sqrt{601}}{6}
$