-16t2+18t+5=0

Solution for -16t2+18t+5=0 equation:

-16t^2+18t+5=0

a = -16; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·(-16)·5
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{161}}{2*-16}=\frac{-18-2\sqrt{161}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{161}}{2*-16}=\frac{-18+2\sqrt{161}}{-32}
$