2p^2+13p+18=0

Solution for 2p^2+13p+18=0 equation:

2p^2+13p+18=0

a = 2; b = 13; c = +18;
Δ = b2-4ac
Δ = 132-4·2·18
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*2}=\frac{-18}{4}
=-4+1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*2}=\frac{-8}{4}
=-2
$