0=-16t^2+128t+71

Solution for 0=-16t^2+128t+71 equation:

0=-16t^2+128t+71

We move all terms to the left:

0-(-16t^2+128t+71)=0

We add all the numbers together, and all the variables

-(-16t^2+128t+71)=0

We get rid of parentheses

16t^2-128t-71=0

a = 16; b = -128; c = -71;
Δ = b2-4ac
Δ = -1282-4·16·(-71)
Δ = 20928
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:

$\sqrt{\Delta}=\sqrt{20928}=\sqrt{64*327}=\sqrt{64}*\sqrt{327}=8\sqrt{327}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-8\sqrt{327}}{2*16}=\frac{128-8\sqrt{327}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+8\sqrt{327}}{2*16}=\frac{128+8\sqrt{327}}{32}
$