p^2+6p-63=0

Solution for p^2+6p-63=0 equation:

p^2+6p-63=0

a = 1; b = 6; c = -63;
Δ = b2-4ac
Δ = 62-4·1·(-63)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12\sqrt{2}}{2*1}=\frac{-6-12\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12\sqrt{2}}{2*1}=\frac{-6+12\sqrt{2}}{2}
$