0.3+p^2+p=1

Solution for 0.3+p^2+p=1 equation:

0.3+p^2+p=1

We move all terms to the left:

0.3+p^2+p-(1)=0

We add all the numbers together, and all the variables

p^2+p-0.7=0

a = 1; b = 1; c = -0.7;
Δ = b2-4ac
Δ = 12-4·1·(-0.7)
Δ = 3.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{3.8}}{2*1}=\frac{-1-\sqrt{3.8}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3.8}}{2*1}=\frac{-1+\sqrt{3.8}}{2}
$