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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

a = -16; b = -240; c = +128;
Δ = b2-4ac
Δ = -2402-4·(-16)·128
Δ = 65792
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:

$\sqrt{\Delta}=\sqrt{65792}=\sqrt{256*257}=\sqrt{256}*\sqrt{257}=16\sqrt{257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-16\sqrt{257}}{2*-16}=\frac{240-16\sqrt{257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+16\sqrt{257}}{2*-16}=\frac{240+16\sqrt{257}}{-32}
$