(p-3)(p-9)+(2p+1)(p-8)=p(2p-23)+31

Solution for (p-3)(p-9)+(2p+1)(p-8)=p(2p-23)+31 equation:

(p-3)(p-9)+(2p+1)(p-8)=p(2p-23)+31

We move all terms to the left:

(p-3)(p-9)+(2p+1)(p-8)-(p(2p-23)+31)=0

We multiply parentheses ..

(+p^2-9p-3p+27)+(2p+1)(p-8)-(p(2p-23)+31)=0


We calculate terms in parentheses: -(p(2p-23)+31), so:

p(2p-23)+31

We multiply parentheses

2p^2-23p+31

Back to the equation:

-(2p^2-23p+31)


We get rid of parentheses

p^2-2p^2-9p-3p+(2p+1)(p-8)+23p+27-31=0

We multiply parentheses ..

p^2-2p^2+(+2p^2-16p+p-8)-9p-3p+23p+27-31=0

We add all the numbers together, and all the variables

-1p^2+(+2p^2-16p+p-8)+11p-4=0

We get rid of parentheses

-1p^2+2p^2-16p+p+11p-8-4=0

We add all the numbers together, and all the variables

p^2-4p-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*1}=\frac{-4}{2}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$