-16t^2+92t+28=0

Solution for -16t^2+92t+28=0 equation:

-16t^2+92t+28=0

a = -16; b = 92; c = +28;
Δ = b2-4ac
Δ = 922-4·(-16)·28
Δ = 10256
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:

$\sqrt{\Delta}=\sqrt{10256}=\sqrt{16*641}=\sqrt{16}*\sqrt{641}=4\sqrt{641}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{641}}{2*-16}=\frac{-92-4\sqrt{641}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{641}}{2*-16}=\frac{-92+4\sqrt{641}}{-32}
$