(p^2-p)=(6p+18)

Solution for (p^2-p)=(6p+18) equation:

(p^2-p)=(6p+18)

We move all terms to the left:

(p^2-p)-((6p+18))=0

We get rid of parentheses

p^2-p-((6p+18))=0


We calculate terms in parentheses: -((6p+18)), so:

(6p+18)

We get rid of parentheses

6p+18

Back to the equation:

-(6p+18)


We add all the numbers together, and all the variables

p^2-1p-(6p+18)=0

We get rid of parentheses

p^2-1p-6p-18=0

We add all the numbers together, and all the variables

p^2-7p-18=0

a = 1; b = -7; c = -18;
Δ = b2-4ac
Δ = -72-4·1·(-18)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*1}=\frac{-4}{2}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*1}=\frac{18}{2}
=9
$