240=-16t^2+128t

Solution for 240=-16t^2+128t equation:

240=-16t^2+128t

We move all terms to the left:

240-(-16t^2+128t)=0

We get rid of parentheses

16t^2-128t+240=0

a = 16; b = -128; c = +240;
Δ = b2-4ac
Δ = -1282-4·16·240
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32}{2*16}=\frac{96}{32}
=3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32}{2*16}=\frac{160}{32}
=5
$