-16t^2-28t+368=0

Solution for -16t^2-28t+368=0 equation:

-16t^2-28t+368=0

a = -16; b = -28; c = +368;
Δ = b2-4ac
Δ = -282-4·(-16)·368
Δ = 24336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{24336}=156$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-156}{2*-16}=\frac{-128}{-32}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+156}{2*-16}=\frac{184}{-32}
=-5+3/4
$