128=16t^2+64t-48

Solution for 128=16t^2+64t-48 equation:

128=16t^2+64t-48

We move all terms to the left:

128-(16t^2+64t-48)=0

We get rid of parentheses

-16t^2-64t+48+128=0

We add all the numbers together, and all the variables

-16t^2-64t+176=0

a = -16; b = -64; c = +176;
Δ = b2-4ac
Δ = -642-4·(-16)·176
Δ = 15360
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:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32\sqrt{15}}{2*-16}=\frac{64-32\sqrt{15}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32\sqrt{15}}{2*-16}=\frac{64+32\sqrt{15}}{-32}
$