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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-32t-46=0

a = 16; b = -32; c = -46;
Δ = b2-4ac
Δ = -322-4·16·(-46)
Δ = 3968
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:

$\sqrt{\Delta}=\sqrt{3968}=\sqrt{64*62}=\sqrt{64}*\sqrt{62}=8\sqrt{62}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{62}}{2*16}=\frac{32-8\sqrt{62}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{62}}{2*16}=\frac{32+8\sqrt{62}}{32}
$