-16t^2+160t-18=0

Solution for -16t^2+160t-18=0 equation:

-16t^2+160t-18=0

a = -16; b = 160; c = -18;
Δ = b2-4ac
Δ = 1602-4·(-16)·(-18)
Δ = 24448
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:

$\sqrt{\Delta}=\sqrt{24448}=\sqrt{64*382}=\sqrt{64}*\sqrt{382}=8\sqrt{382}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{382}}{2*-16}=\frac{-160-8\sqrt{382}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{382}}{2*-16}=\frac{-160+8\sqrt{382}}{-32}
$