4=-16t^2+127

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

4=-16t^2+127

We move all terms to the left:

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

We get rid of parentheses

16t^2-127+4=0

We add all the numbers together, and all the variables

16t^2-123=0

a = 16; b = 0; c = -123;
Δ = b2-4ac
Δ = 02-4·16·(-123)
Δ = 7872
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:

$\sqrt{\Delta}=\sqrt{7872}=\sqrt{64*123}=\sqrt{64}*\sqrt{123}=8\sqrt{123}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{123}}{2*16}=\frac{0-8\sqrt{123}}{32}
=-\frac{8\sqrt{123}}{32}
=-\frac{\sqrt{123}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{123}}{2*16}=\frac{0+8\sqrt{123}}{32}
=\frac{8\sqrt{123}}{32}
=\frac{\sqrt{123}}{4}
$