160=92t+16t^2

Solution for 160=92t+16t^2 equation:

160=92t+16t^2

We move all terms to the left:

160-(92t+16t^2)=0

We get rid of parentheses

-16t^2-92t+160=0

a = -16; b = -92; c = +160;
Δ = b2-4ac
Δ = -922-4·(-16)·160
Δ = 18704
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:

$\sqrt{\Delta}=\sqrt{18704}=\sqrt{16*1169}=\sqrt{16}*\sqrt{1169}=4\sqrt{1169}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-4\sqrt{1169}}{2*-16}=\frac{92-4\sqrt{1169}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+4\sqrt{1169}}{2*-16}=\frac{92+4\sqrt{1169}}{-32}
$