-16t^2+144t-128=0

Solution for -16t^2+144t-128=0 equation:

-16t^2+144t-128=0

a = -16; b = 144; c = -128;
Δ = b2-4ac
Δ = 1442-4·(-16)·(-128)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-112}{2*-16}=\frac{-256}{-32}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+112}{2*-16}=\frac{-32}{-32}
=1
$