32t=128+16t-16t^2

Solution for 32t=128+16t-16t^2 equation:

32t=128+16t-16t^2

We move all terms to the left:

32t-(128+16t-16t^2)=0

We get rid of parentheses

16t^2-16t+32t-128=0

We add all the numbers together, and all the variables

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}
$