2=4+65t-16t^2

Solution for 2=4+65t-16t^2 equation:

2=4+65t-16t^2

We move all terms to the left:

2-(4+65t-16t^2)=0

We get rid of parentheses

16t^2-65t-4+2=0

We add all the numbers together, and all the variables

16t^2-65t-2=0

a = 16; b = -65; c = -2;
Δ = b2-4ac
Δ = -652-4·16·(-2)
Δ = 4353
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-\sqrt{4353}}{2*16}=\frac{65-\sqrt{4353}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+\sqrt{4353}}{2*16}=\frac{65+\sqrt{4353}}{32}
$