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

Solution for -16t^2+18t-5=0 equation:

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

a = -16; b = 18; c = -5;
Δ = b2-4ac
Δ = 182-4·(-16)·(-5)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*-16}=\frac{-20}{-32}
=5/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*-16}=\frac{-16}{-32}
=1/2
$