1308=16t^2

Solution for 1308=16t^2 equation:

1308=16t^2

We move all terms to the left:

1308-(16t^2)=0

a = -16; b = 0; c = +1308;
Δ = b2-4ac
Δ = 02-4·(-16)·1308
Δ = 83712
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:

$\sqrt{\Delta}=\sqrt{83712}=\sqrt{256*327}=\sqrt{256}*\sqrt{327}=16\sqrt{327}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{327}}{2*-16}=\frac{0-16\sqrt{327}}{-32}
=-\frac{16\sqrt{327}}{-32}
=-\frac{\sqrt{327}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{327}}{2*-16}=\frac{0+16\sqrt{327}}{-32}
=\frac{16\sqrt{327}}{-32}
=\frac{\sqrt{327}}{-2}
$