112t-16t^2=160

Solution for 112t-16t^2=160 equation:

112t-16t^2=160

We move all terms to the left:

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