112t-16t^2-96=0

Solution for 112t-16t^2-96=0 equation:

112t-16t^2-96=0

a = -16; b = 112; c = -96;
Δ = b2-4ac
Δ = 1122-4·(-16)·(-96)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-80}{2*-16}=\frac{-192}{-32}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+80}{2*-16}=\frac{-32}{-32}
=1
$