1=-16t2+32t+48

Solution for 1=-16t2+32t+48 equation:

1=-16t^2+32t+48

We move all terms to the left:

1-(-16t^2+32t+48)=0

We get rid of parentheses

16t^2-32t-48+1=0

We add all the numbers together, and all the variables

16t^2-32t-47=0

a = 16; b = -32; c = -47;
Δ = b2-4ac
Δ = -322-4·16·(-47)
Δ = 4032
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:

$\sqrt{\Delta}=\sqrt{4032}=\sqrt{576*7}=\sqrt{576}*\sqrt{7}=24\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24\sqrt{7}}{2*16}=\frac{32-24\sqrt{7}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24\sqrt{7}}{2*16}=\frac{32+24\sqrt{7}}{32}
$