T=128+16t-16t^2

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

=128+16T-16T^2

We move all terms to the left:

-(128+16T-16T^2)=0

We get rid of parentheses

16T^2-16T-128=0

a = 16; b = -16; c = -128;
Δ = b2-4ac
Δ = -162-4·16·(-128)
Δ = 8448
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:

$\sqrt{\Delta}=\sqrt{8448}=\sqrt{256*33}=\sqrt{256}*\sqrt{33}=16\sqrt{33}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{33}}{2*16}=\frac{16-16\sqrt{33}}{32}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{33}}{2*16}=\frac{16+16\sqrt{33}}{32}
$