48=32t+16t^2

Solution for 48=32t+16t^2 equation:

48=32t+16t^2

We move all terms to the left:

48-(32t+16t^2)=0

We get rid of parentheses

-16t^2-32t+48=0

a = -16; b = -32; c = +48;
Δ = b2-4ac
Δ = -322-4·(-16)·48
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-64}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+64}{2*-16}=\frac{96}{-32}
=-3
$