12t^2-24t-108=0

Solution for 12t^2-24t-108=0 equation:

12t^2-24t-108=0

a = 12; b = -24; c = -108;
Δ = b2-4ac
Δ = -242-4·12·(-108)
Δ = 5760
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:

$\sqrt{\Delta}=\sqrt{5760}=\sqrt{576*10}=\sqrt{576}*\sqrt{10}=24\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24\sqrt{10}}{2*12}=\frac{24-24\sqrt{10}}{24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24\sqrt{10}}{2*12}=\frac{24+24\sqrt{10}}{24}
$