160-24t-16t^2=0

Solution for 160-24t-16t^2=0 equation:

160-24t-16t^2=0

a = -16; b = -24; c = +160;
Δ = b2-4ac
Δ = -242-4·(-16)·160
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-104}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+104}{2*-16}=\frac{128}{-32}
=-4
$