112-96t-16t^2=0

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

112-96t-16t^2=0

a = -16; b = -96; c = +112;
Δ = b2-4ac
Δ = -962-4·(-16)·112
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-128}{2*-16}=\frac{-32}{-32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+128}{2*-16}=\frac{224}{-32}
=-7
$