30=-3t^2+120t

Solution for 30=-3t^2+120t equation:

30=-3t^2+120t

We move all terms to the left:

30-(-3t^2+120t)=0

We get rid of parentheses

3t^2-120t+30=0

a = 3; b = -120; c = +30;
Δ = b2-4ac
Δ = -1202-4·3·30
Δ = 14040
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14040}=\sqrt{36*390}=\sqrt{36}*\sqrt{390}=6\sqrt{390}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-6\sqrt{390}}{2*3}=\frac{120-6\sqrt{390}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+6\sqrt{390}}{2*3}=\frac{120+6\sqrt{390}}{6}
$