0=-16t^2+50t+120

Solution for 0=-16t^2+50t+120 equation:

0=-16t^2+50t+120

We move all terms to the left:

0-(-16t^2+50t+120)=0

We add all the numbers together, and all the variables

-(-16t^2+50t+120)=0

We get rid of parentheses

16t^2-50t-120=0

a = 16; b = -50; c = -120;
Δ = b2-4ac
Δ = -502-4·16·(-120)
Δ = 10180
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:

$\sqrt{\Delta}=\sqrt{10180}=\sqrt{4*2545}=\sqrt{4}*\sqrt{2545}=2\sqrt{2545}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{2545}}{2*16}=\frac{50-2\sqrt{2545}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{2545}}{2*16}=\frac{50+2\sqrt{2545}}{32}
$