p^2+14p+18=0

Solution for p^2+14p+18=0 equation:

p^2+14p+18=0

a = 1; b = 14; c = +18;
Δ = b2-4ac
Δ = 142-4·1·18
Δ = 124
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:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{31}}{2*1}=\frac{-14-2\sqrt{31}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{31}}{2*1}=\frac{-14+2\sqrt{31}}{2}
$