0=-16t^2+30t+120

Solution for 0=-16t^2+30t+120 equation:

0=-16t^2+30t+120

We move all terms to the left:

0-(-16t^2+30t+120)=0

We add all the numbers together, and all the variables

-(-16t^2+30t+120)=0

We get rid of parentheses

16t^2-30t-120=0

a = 16; b = -30; c = -120;
Δ = b2-4ac
Δ = -302-4·16·(-120)
Δ = 8580
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:

$\sqrt{\Delta}=\sqrt{8580}=\sqrt{4*2145}=\sqrt{4}*\sqrt{2145}=2\sqrt{2145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{2145}}{2*16}=\frac{30-2\sqrt{2145}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{2145}}{2*16}=\frac{30+2\sqrt{2145}}{32}
$