-16t^2-32t+235=0

Solution for -16t^2-32t+235=0 equation:

-16t^2-32t+235=0

a = -16; b = -32; c = +235;
Δ = b2-4ac
Δ = -322-4·(-16)·235
Δ = 16064
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:

$\sqrt{\Delta}=\sqrt{16064}=\sqrt{64*251}=\sqrt{64}*\sqrt{251}=8\sqrt{251}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{251}}{2*-16}=\frac{32-8\sqrt{251}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{251}}{2*-16}=\frac{32+8\sqrt{251}}{-32}
$