120=22t+2t^2

Solution for 120=22t+2t^2 equation:

120=22t+2t^2

We move all terms to the left:

120-(22t+2t^2)=0

We get rid of parentheses

-2t^2-22t+120=0

a = -2; b = -22; c = +120;
Δ = b2-4ac
Δ = -222-4·(-2)·120
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-38}{2*-2}=\frac{-16}{-4}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+38}{2*-2}=\frac{60}{-4}
=-15
$