16(t-2)(t-5)=0

Solution for 16(t-2)(t-5)=0 equation:

16(t-2)(t-5)=0

We multiply parentheses ..

16(+t^2-5t-2t+10)=0

We multiply parentheses

16t^2-80t-32t+160=0

We add all the numbers together, and all the variables

16t^2-112t+160=0

a = 16; b = -112; c = +160;
Δ = b2-4ac
Δ = -1122-4·16·160
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-48}{2*16}=\frac{64}{32}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+48}{2*16}=\frac{160}{32}
=5
$