2=-16t^2+48t+4

Solution for 2=-16t^2+48t+4 equation:

2=-16t^2+48t+4

We move all terms to the left:

2-(-16t^2+48t+4)=0

We get rid of parentheses

16t^2-48t-4+2=0

We add all the numbers together, and all the variables

16t^2-48t-2=0

a = 16; b = -48; c = -2;
Δ = b2-4ac
Δ = -482-4·16·(-2)
Δ = 2432
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:

$\sqrt{\Delta}=\sqrt{2432}=\sqrt{64*38}=\sqrt{64}*\sqrt{38}=8\sqrt{38}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{38}}{2*16}=\frac{48-8\sqrt{38}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{38}}{2*16}=\frac{48+8\sqrt{38}}{32}
$