4000=120t+0.9t^2

Solution for 4000=120t+0.9t^2 equation:

4000=120t+0.9t^2

We move all terms to the left:

4000-(120t+0.9t^2)=0

We get rid of parentheses

-0.9t^2-120t+4000=0

a = -0.9; b = -120; c = +4000;
Δ = b2-4ac
Δ = -1202-4·(-0.9)·4000
Δ = 28800
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:

$\sqrt{\Delta}=\sqrt{28800}=\sqrt{14400*2}=\sqrt{14400}*\sqrt{2}=120\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120\sqrt{2}}{2*-0.9}=\frac{120-120\sqrt{2}}{-1.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120\sqrt{2}}{2*-0.9}=\frac{120+120\sqrt{2}}{-1.8}
$