6p^2+18p+12=0

Solution for 6p^2+18p+12=0 equation:

6p^2+18p+12=0

a = 6; b = 18; c = +12;
Δ = b2-4ac
Δ = 182-4·6·12
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*6}=\frac{-24}{12}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*6}=\frac{-12}{12}
=-1
$