28=-16t^2+1444

Solution for 28=-16t^2+1444 equation:

28=-16t^2+1444

We move all terms to the left:

28-(-16t^2+1444)=0

We get rid of parentheses

16t^2-1444+28=0

We add all the numbers together, and all the variables

16t^2-1416=0

a = 16; b = 0; c = -1416;
Δ = b2-4ac
Δ = 02-4·16·(-1416)
Δ = 90624
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:

$\sqrt{\Delta}=\sqrt{90624}=\sqrt{256*354}=\sqrt{256}*\sqrt{354}=16\sqrt{354}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{354}}{2*16}=\frac{0-16\sqrt{354}}{32}
=-\frac{16\sqrt{354}}{32}
=-\frac{\sqrt{354}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{354}}{2*16}=\frac{0+16\sqrt{354}}{32}
=\frac{16\sqrt{354}}{32}
=\frac{\sqrt{354}}{2}
$