127=-16t^2+100t

Solution for 127=-16t^2+100t equation:

127=-16t^2+100t

We move all terms to the left:

127-(-16t^2+100t)=0

We get rid of parentheses

16t^2-100t+127=0

a = 16; b = -100; c = +127;
Δ = b2-4ac
Δ = -1002-4·16·127
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-12\sqrt{13}}{2*16}=\frac{100-12\sqrt{13}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+12\sqrt{13}}{2*16}=\frac{100+12\sqrt{13}}{32}
$