-16t^2+116t-28=0

Solution for -16t^2+116t-28=0 equation:

-16t^2+116t-28=0

a = -16; b = 116; c = -28;
Δ = b2-4ac
Δ = 1162-4·(-16)·(-28)
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(116)-108}{2*-16}=\frac{-224}{-32}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(116)+108}{2*-16}=\frac{-8}{-32}
=1/4
$