2+44t-16t^2=0

Solution for 2+44t-16t^2=0 equation:

2+44t-16t^2=0

a = -16; b = 44; c = +2;
Δ = b2-4ac
Δ = 442-4·(-16)·2
Δ = 2064
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:

$\sqrt{\Delta}=\sqrt{2064}=\sqrt{16*129}=\sqrt{16}*\sqrt{129}=4\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{129}}{2*-16}=\frac{-44-4\sqrt{129}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{129}}{2*-16}=\frac{-44+4\sqrt{129}}{-32}
$