-16t^2+113t+65=0

Solution for -16t^2+113t+65=0 equation:

-16t^2+113t+65=0

a = -16; b = 113; c = +65;
Δ = b2-4ac
Δ = 1132-4·(-16)·65
Δ = 16929
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:

$\sqrt{\Delta}=\sqrt{16929}=\sqrt{81*209}=\sqrt{81}*\sqrt{209}=9\sqrt{209}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(113)-9\sqrt{209}}{2*-16}=\frac{-113-9\sqrt{209}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(113)+9\sqrt{209}}{2*-16}=\frac{-113+9\sqrt{209}}{-32}
$