p^2+2p-55=0

Solution for p^2+2p-55=0 equation:

p^2+2p-55=0

a = 1; b = 2; c = -55;
Δ = b2-4ac
Δ = 22-4·1·(-55)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{14}}{2*1}=\frac{-2-4\sqrt{14}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{14}}{2*1}=\frac{-2+4\sqrt{14}}{2}
$