3p^2+19p+16=0

Solution for 3p^2+19p+16=0 equation:

3p^2+19p+16=0

a = 3; b = 19; c = +16;
Δ = b2-4ac
Δ = 192-4·3·16
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*3}=\frac{-32}{6}
=-5+1/3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*3}=\frac{-6}{6}
=-1
$