t^2-24t-132=0

Solution for t^2-24t-132=0 equation:

t^2-24t-132=0

a = 1; b = -24; c = -132;
Δ = b2-4ac
Δ = -242-4·1·(-132)
Δ = 1104
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:

$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{69}}{2*1}=\frac{24-4\sqrt{69}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{69}}{2*1}=\frac{24+4\sqrt{69}}{2}
$