48=16t^2+16t

Solution for 48=16t^2+16t equation:

48=16t^2+16t

We move all terms to the left:

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

We get rid of parentheses

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

a = -16; b = -16; c = +48;
Δ = b2-4ac
Δ = -162-4·(-16)·48
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{13}}{2*-16}=\frac{16-16\sqrt{13}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{13}}{2*-16}=\frac{16+16\sqrt{13}}{-32}
$