1145=16t^2

Solution for 1145=16t^2 equation:

1145=16t^2

We move all terms to the left:

1145-(16t^2)=0

a = -16; b = 0; c = +1145;
Δ = b2-4ac
Δ = 02-4·(-16)·1145
Δ = 73280
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:

$\sqrt{\Delta}=\sqrt{73280}=\sqrt{64*1145}=\sqrt{64}*\sqrt{1145}=8\sqrt{1145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{1145}}{2*-16}=\frac{0-8\sqrt{1145}}{-32}
=-\frac{8\sqrt{1145}}{-32}
=-\frac{\sqrt{1145}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{1145}}{2*-16}=\frac{0+8\sqrt{1145}}{-32}
=\frac{8\sqrt{1145}}{-32}
=\frac{\sqrt{1145}}{-4}
$