-16t^2+112t+48=0

Solution for -16t^2+112t+48=0 equation:

-16t^2+112t+48=0

a = -16; b = 112; c = +48;
Δ = b2-4ac
Δ = 1122-4·(-16)·48
Δ = 15616
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:

$\sqrt{\Delta}=\sqrt{15616}=\sqrt{256*61}=\sqrt{256}*\sqrt{61}=16\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-16\sqrt{61}}{2*-16}=\frac{-112-16\sqrt{61}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+16\sqrt{61}}{2*-16}=\frac{-112+16\sqrt{61}}{-32}
$