48=16t^2+80t

Solution for 48=16t^2+80t equation:

48=16t^2+80t

We move all terms to the left:

48-(16t^2+80t)=0

We get rid of parentheses

-16t^2-80t+48=0

a = -16; b = -80; c = +48;
Δ = b2-4ac
Δ = -802-4·(-16)·48
Δ = 9472
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:

$\sqrt{\Delta}=\sqrt{9472}=\sqrt{256*37}=\sqrt{256}*\sqrt{37}=16\sqrt{37}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-16\sqrt{37}}{2*-16}=\frac{80-16\sqrt{37}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+16\sqrt{37}}{2*-16}=\frac{80+16\sqrt{37}}{-32}
$