12p^2+12p-18=0

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

12p^2+12p-18=0

a = 12; b = 12; c = -18;
Δ = b2-4ac
Δ = 122-4·12·(-18)
Δ = 1008
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{7}}{2*12}=\frac{-12-12\sqrt{7}}{24}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{7}}{2*12}=\frac{-12+12\sqrt{7}}{24}
$