p^2+10p-10=p

Solution for p^2+10p-10=p equation:

p^2+10p-10=p

We move all terms to the left:

p^2+10p-10-(p)=0

We add all the numbers together, and all the variables

p^2+9p-10=0

a = 1; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·1·(-10)
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{-(9)-11}{2*1}=\frac{-20}{2}
=-10
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*1}=\frac{2}{2}
=1
$