-16t^2+54t-18=0

Solution for -16t^2+54t-18=0 equation:

-16t^2+54t-18=0

a = -16; b = 54; c = -18;
Δ = b2-4ac
Δ = 542-4·(-16)·(-18)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-42}{2*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+42}{2*-16}=\frac{-12}{-32}
=3/8
$