4=-16t^2+127t

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

4=-16t^2+127t

We move all terms to the left:

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

We get rid of parentheses

16t^2-127t+4=0

a = 16; b = -127; c = +4;
Δ = b2-4ac
Δ = -1272-4·16·4
Δ = 15873
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{15873}}{2*16}=\frac{127-\sqrt{15873}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{15873}}{2*16}=\frac{127+\sqrt{15873}}{32}
$