120=-t^2+50t

Solution for 120=-t^2+50t equation:

120=-t^2+50t

We move all terms to the left:

120-(-t^2+50t)=0

We get rid of parentheses

t^2-50t+120=0

a = 1; b = -50; c = +120;
Δ = b2-4ac
Δ = -502-4·1·120
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{505}}{2*1}=\frac{50-2\sqrt{505}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{505}}{2*1}=\frac{50+2\sqrt{505}}{2}
$