16t^2+120t+80=0

Solution for 16t^2+120t+80=0 equation:

16t^2+120t+80=0

a = 16; b = 120; c = +80;
Δ = b2-4ac
Δ = 1202-4·16·80
Δ = 9280
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:

$\sqrt{\Delta}=\sqrt{9280}=\sqrt{64*145}=\sqrt{64}*\sqrt{145}=8\sqrt{145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-8\sqrt{145}}{2*16}=\frac{-120-8\sqrt{145}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+8\sqrt{145}}{2*16}=\frac{-120+8\sqrt{145}}{32}
$