2+30t-16t^2=0

Solution for 2+30t-16t^2=0 equation:

2+30t-16t^2=0

a = -16; b = 30; c = +2;
Δ = b2-4ac
Δ = 302-4·(-16)·2
Δ = 1028
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:

$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{257}}{2*-16}=\frac{-30-2\sqrt{257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{257}}{2*-16}=\frac{-30+2\sqrt{257}}{-32}
$