12t^2-48t-48=0

Solution for 12t^2-48t-48=0 equation:

12t^2-48t-48=0

a = 12; b = -48; c = -48;
Δ = b2-4ac
Δ = -482-4·12·(-48)
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48\sqrt{2}}{2*12}=\frac{48-48\sqrt{2}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48\sqrt{2}}{2*12}=\frac{48+48\sqrt{2}}{24}
$