16t^2+192t+320=0

Solution for 16t^2+192t+320=0 equation:

16t^2+192t+320=0

a = 16; b = 192; c = +320;
Δ = b2-4ac
Δ = 1922-4·16·320
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-128}{2*16}=\frac{-320}{32}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+128}{2*16}=\frac{-64}{32}
=-2
$