9p^2+19p+2=0

Solution for 9p^2+19p+2=0 equation:

9p^2+19p+2=0

a = 9; b = 19; c = +2;
Δ = b2-4ac
Δ = 192-4·9·2
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*9}=\frac{-36}{18}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*9}=\frac{-2}{18}
=-1/9
$