t=-16t2-18t+120

Solution for t=-16t2-18t+120 equation:

t=-16t^2-18t+120

We move all terms to the left:

t-(-16t^2-18t+120)=0

We get rid of parentheses

16t^2+18t+t-120=0

We add all the numbers together, and all the variables

16t^2+19t-120=0

a = 16; b = 19; c = -120;
Δ = b2-4ac
Δ = 192-4·16·(-120)
Δ = 8041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{8041}}{2*16}=\frac{-19-\sqrt{8041}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{8041}}{2*16}=\frac{-19+\sqrt{8041}}{32}
$