-16t^2+112t+42=0

Solution for -16t^2+112t+42=0 equation:

-16t^2+112t+42=0

a = -16; b = 112; c = +42;
Δ = b2-4ac
Δ = 1122-4·(-16)·42
Δ = 15232
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:

$\sqrt{\Delta}=\sqrt{15232}=\sqrt{64*238}=\sqrt{64}*\sqrt{238}=8\sqrt{238}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-8\sqrt{238}}{2*-16}=\frac{-112-8\sqrt{238}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+8\sqrt{238}}{2*-16}=\frac{-112+8\sqrt{238}}{-32}
$