-16t^2+248=0

Solution for -16t^2+248=0 equation:

-16t^2+248=0

a = -16; b = 0; c = +248;
Δ = b2-4ac
Δ = 02-4·(-16)·248
Δ = 15872
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:

$\sqrt{\Delta}=\sqrt{15872}=\sqrt{256*62}=\sqrt{256}*\sqrt{62}=16\sqrt{62}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{62}}{2*-16}=\frac{0-16\sqrt{62}}{-32}
=-\frac{16\sqrt{62}}{-32}
=-\frac{\sqrt{62}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{62}}{2*-16}=\frac{0+16\sqrt{62}}{-32}
=\frac{16\sqrt{62}}{-32}
=\frac{\sqrt{62}}{-2}
$