-p^2+12p-24=0

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

-p^2+12p-24=0

We add all the numbers together, and all the variables

-1p^2+12p-24=0

a = -1; b = 12; c = -24;
Δ = b2-4ac
Δ = 122-4·(-1)·(-24)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{3}}{2*-1}=\frac{-12-4\sqrt{3}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*-1}=\frac{-12+4\sqrt{3}}{-2}
$