t^2+12t-24=0

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

t^2+12t-24=0

a = 1; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·1·(-24)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*1}=\frac{-12-4\sqrt{15}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*1}=\frac{-12+4\sqrt{15}}{2}
$