24t2+192t+360=0

Solution for 24t2+192t+360=0 equation:

24t^2+192t+360=0

a = 24; b = 192; c = +360;
Δ = b2-4ac
Δ = 1922-4·24·360
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{-(192)-48}{2*24}=\frac{-240}{48}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+48}{2*24}=\frac{-144}{48}
=-3
$