t=-16t^2+96t+128

Solution for t=-16t^2+96t+128 equation:

t=-16t^2+96t+128

We move all terms to the left:

t-(-16t^2+96t+128)=0

We get rid of parentheses

16t^2-96t+t-128=0

We add all the numbers together, and all the variables

16t^2-95t-128=0

a = 16; b = -95; c = -128;
Δ = b2-4ac
Δ = -952-4·16·(-128)
Δ = 17217
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:

$\sqrt{\Delta}=\sqrt{17217}=\sqrt{9*1913}=\sqrt{9}*\sqrt{1913}=3\sqrt{1913}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-3\sqrt{1913}}{2*16}=\frac{95-3\sqrt{1913}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+3\sqrt{1913}}{2*16}=\frac{95+3\sqrt{1913}}{32}
$