t^2-12t-96=0

Solution for t^2-12t-96=0 equation:

t^2-12t-96=0

a = 1; b = -12; c = -96;
Δ = b2-4ac
Δ = -122-4·1·(-96)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{33}}{2*1}=\frac{12-4\sqrt{33}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{33}}{2*1}=\frac{12+4\sqrt{33}}{2}
$