-16t^2+7t+32=0

Solution for -16t^2+7t+32=0 equation:

-16t^2+7t+32=0

a = -16; b = 7; c = +32;
Δ = b2-4ac
Δ = 72-4·(-16)·32
Δ = 2097
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:

$\sqrt{\Delta}=\sqrt{2097}=\sqrt{9*233}=\sqrt{9}*\sqrt{233}=3\sqrt{233}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{233}}{2*-16}=\frac{-7-3\sqrt{233}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{233}}{2*-16}=\frac{-7+3\sqrt{233}}{-32}
$