p^2+18p+2=0

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

p^2+18p+2=0

a = 1; b = 18; c = +2;
Δ = b2-4ac
Δ = 182-4·1·2
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{79}}{2*1}=\frac{-18-2\sqrt{79}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{79}}{2*1}=\frac{-18+2\sqrt{79}}{2}
$