-8t^2+48t+1728=0

Solution for -8t^2+48t+1728=0 equation:

-8t^2+48t+1728=0

a = -8; b = 48; c = +1728;
Δ = b2-4ac
Δ = 482-4·(-8)·1728
Δ = 57600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57600}=240$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-240}{2*-8}=\frac{-288}{-16}
=+18
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+240}{2*-8}=\frac{192}{-16}
=-12
$