1148=16t^2

Solution for 1148=16t^2 equation:

1148=16t^2

We move all terms to the left:

1148-(16t^2)=0

a = -16; b = 0; c = +1148;
Δ = b2-4ac
Δ = 02-4·(-16)·1148
Δ = 73472
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:

$\sqrt{\Delta}=\sqrt{73472}=\sqrt{256*287}=\sqrt{256}*\sqrt{287}=16\sqrt{287}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{287}}{2*-16}=\frac{0-16\sqrt{287}}{-32}
=-\frac{16\sqrt{287}}{-32}
=-\frac{\sqrt{287}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{287}}{2*-16}=\frac{0+16\sqrt{287}}{-32}
=\frac{16\sqrt{287}}{-32}
=\frac{\sqrt{287}}{-2}
$