2=-16t^2+32t+5

Solution for 2=-16t^2+32t+5 equation:

2=-16t^2+32t+5

We move all terms to the left:

2-(-16t^2+32t+5)=0

We get rid of parentheses

16t^2-32t-5+2=0

We add all the numbers together, and all the variables

16t^2-32t-3=0

a = 16; b = -32; c = -3;
Δ = b2-4ac
Δ = -322-4·16·(-3)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{19}}{2*16}=\frac{32-8\sqrt{19}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{19}}{2*16}=\frac{32+8\sqrt{19}}{32}
$