p^2+12p-97=-8

Solution for p^2+12p-97=-8 equation:

p^2+12p-97=-8

We move all terms to the left:

p^2+12p-97-(-8)=0

We add all the numbers together, and all the variables

p^2+12p-89=0

a = 1; b = 12; c = -89;
Δ = b2-4ac
Δ = 122-4·1·(-89)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10\sqrt{5}}{2*1}=\frac{-12-10\sqrt{5}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10\sqrt{5}}{2*1}=\frac{-12+10\sqrt{5}}{2}
$