127=-16t^2+127t+4

Solution for 127=-16t^2+127t+4 equation:

127=-16t^2+127t+4

We move all terms to the left:

127-(-16t^2+127t+4)=0

We get rid of parentheses

16t^2-127t-4+127=0

We add all the numbers together, and all the variables

16t^2-127t+123=0

a = 16; b = -127; c = +123;
Δ = b2-4ac
Δ = -1272-4·16·123
Δ = 8257
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-127)-\sqrt{8257}}{2*16}=\frac{127-\sqrt{8257}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{8257}}{2*16}=\frac{127+\sqrt{8257}}{32}
$