p2+18p+80=0

Solution for p2+18p+80=0 equation:

p2+18p+80=0

We add all the numbers together, and all the variables

p^2+18p+80=0

a = 1; b = 18; c = +80;
Δ = b2-4ac
Δ = 182-4·1·80
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*1}=\frac{-20}{2}
=-10
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*1}=\frac{-16}{2}
=-8
$