t=-16t^2+128t+3500

Solution for t=-16t^2+128t+3500 equation:

t=-16t^2+128t+3500

We move all terms to the left:

t-(-16t^2+128t+3500)=0

We get rid of parentheses

16t^2-128t+t-3500=0

We add all the numbers together, and all the variables

16t^2-127t-3500=0

a = 16; b = -127; c = -3500;
Δ = b2-4ac
Δ = -1272-4·16·(-3500)
Δ = 240129
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:

$\sqrt{\Delta}=\sqrt{240129}=\sqrt{9*26681}=\sqrt{9}*\sqrt{26681}=3\sqrt{26681}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-127)-3\sqrt{26681}}{2*16}=\frac{127-3\sqrt{26681}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+3\sqrt{26681}}{2*16}=\frac{127+3\sqrt{26681}}{32}
$