440=16t^2

Solution for 440=16t^2 equation:

440=16t^2

We move all terms to the left:

440-(16t^2)=0

a = -16; b = 0; c = +440;
Δ = b2-4ac
Δ = 02-4·(-16)·440
Δ = 28160
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:

$\sqrt{\Delta}=\sqrt{28160}=\sqrt{256*110}=\sqrt{256}*\sqrt{110}=16\sqrt{110}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{110}}{2*-16}=\frac{0-16\sqrt{110}}{-32}
=-\frac{16\sqrt{110}}{-32}
=-\frac{\sqrt{110}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{110}}{2*-16}=\frac{0+16\sqrt{110}}{-32}
=\frac{16\sqrt{110}}{-32}
=\frac{\sqrt{110}}{-2}
$