-16t^2+96t+112=100

Solution for -16t^2+96t+112=100 equation:

-16t^2+96t+112=100

We move all terms to the left:

-16t^2+96t+112-(100)=0

We add all the numbers together, and all the variables

-16t^2+96t+12=0

a = -16; b = 96; c = +12;
Δ = b2-4ac
Δ = 962-4·(-16)·12
Δ = 9984
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:

$\sqrt{\Delta}=\sqrt{9984}=\sqrt{256*39}=\sqrt{256}*\sqrt{39}=16\sqrt{39}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{39}}{2*-16}=\frac{-96-16\sqrt{39}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{39}}{2*-16}=\frac{-96+16\sqrt{39}}{-32}
$