(5/5-p)-(p^2/5-p)=-2

Solution for (5/5-p)-(p^2/5-p)=-2 equation:

(5/5-p)-(p^2/5-p)=-2

We move all terms to the left:

(5/5-p)-(p^2/5-p)-(-2)=0


Domain of the equation: 5-p)!=0

We move all terms containing p to the left, all other terms to the right

-p)!=-5

p!=-5/1

p!=-5

p∈R


We add all the numbers together, and all the variables

(-1p+1)-(p^2/5-p)-(-2)=0

We add all the numbers together, and all the variables

(-1p+1)-(p^2/5-p)+2=0

We get rid of parentheses

-p^2/5-1p+p+1+2=0

We multiply all the terms by the denominator


-p^2-1p*5+p*5+1*5+2*5=0

We add all the numbers together, and all the variables

-1p^2-1p*5+p*5+15=0

Wy multiply elements

-1p^2-5p+5p+15=0

We add all the numbers together, and all the variables

-1p^2+15=0

a = -1; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-1)·15
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*-1}=\frac{0-2\sqrt{15}}{-2}
=-\frac{2\sqrt{15}}{-2}
=-\frac{\sqrt{15}}{-1}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*-1}=\frac{0+2\sqrt{15}}{-2}
=\frac{2\sqrt{15}}{-2}
=\frac{\sqrt{15}}{-1}
$