0=160-16(t)^2

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

0=160-16(t)^2

We move all terms to the left:

0-(160-16(t)^2)=0

We add all the numbers together, and all the variables

-(160-16t^2)=0

We get rid of parentheses

16t^2-160=0

a = 16; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·16·(-160)
Δ = 10240
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:

$\sqrt{\Delta}=\sqrt{10240}=\sqrt{1024*10}=\sqrt{1024}*\sqrt{10}=32\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{10}}{2*16}=\frac{0-32\sqrt{10}}{32}
=-\frac{32\sqrt{10}}{32}
=-\sqrt{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{10}}{2*16}=\frac{0+32\sqrt{10}}{32}
=\frac{32\sqrt{10}}{32}
=\sqrt{10}
$