1=-p(1-3p)

Solution for 1=-p(1-3p) equation:

1=-p(1-3p)

We move all terms to the left:

1-(-p(1-3p))=0

We add all the numbers together, and all the variables

-(-p(-3p+1))+1=0


We calculate terms in parentheses: -(-p(-3p+1)), so:

-p(-3p+1)

We multiply parentheses

3p^2-p

We add all the numbers together, and all the variables

3p^2-1p

Back to the equation:

-(3p^2-1p)


We get rid of parentheses

-3p^2+1p+1=0

We add all the numbers together, and all the variables

-3p^2+p+1=0

a = -3; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-3)·1
Δ = 13
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{13}}{2*-3}=\frac{-1-\sqrt{13}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-3}=\frac{-1+\sqrt{13}}{-6}
$