192(t)=160+48t-16t^2

Solution for 192(t)=160+48t-16t^2 equation:

192(t)=160+48t-16t^2

We move all terms to the left:

192(t)-(160+48t-16t^2)=0

We get rid of parentheses

16t^2-48t+192t-160=0

We add all the numbers together, and all the variables

16t^2+144t-160=0

a = 16; b = 144; c = -160;
Δ = b2-4ac
Δ = 1442-4·16·(-160)
Δ = 30976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{30976}=176$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-176}{2*16}=\frac{-320}{32}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+176}{2*16}=\frac{32}{32}
=1
$