-16t^2+18t-4=0

Solution for -16t^2+18t-4=0 equation:

-16t^2+18t-4=0

a = -16; b = 18; c = -4;
Δ = b2-4ac
Δ = 182-4·(-16)·(-4)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{17}}{2*-16}=\frac{-18-2\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{17}}{2*-16}=\frac{-18+2\sqrt{17}}{-32}
$