0=-16t^2+1972

Solution for 0=-16t^2+1972 equation:

0=-16t^2+1972

We move all terms to the left:

0-(-16t^2+1972)=0

We add all the numbers together, and all the variables

-(-16t^2+1972)=0

We get rid of parentheses

16t^2-1972=0

a = 16; b = 0; c = -1972;
Δ = b2-4ac
Δ = 02-4·16·(-1972)
Δ = 126208
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:

$\sqrt{\Delta}=\sqrt{126208}=\sqrt{256*493}=\sqrt{256}*\sqrt{493}=16\sqrt{493}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{493}}{2*16}=\frac{0-16\sqrt{493}}{32}
=-\frac{16\sqrt{493}}{32}
=-\frac{\sqrt{493}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{493}}{2*16}=\frac{0+16\sqrt{493}}{32}
=\frac{16\sqrt{493}}{32}
=\frac{\sqrt{493}}{2}
$