p^2+16p+18=0

Solution for p^2+16p+18=0 equation:

p^2+16p+18=0

a = 1; b = 16; c = +18;
Δ = b2-4ac
Δ = 162-4·1·18
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{46}}{2*1}=\frac{-16-2\sqrt{46}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{46}}{2*1}=\frac{-16+2\sqrt{46}}{2}
$