(p2-3)-2,p=1

Solution for (p2-3)-2,p=1 equation:

(p2-3)-2.p=1

We move all terms to the left:

(p2-3)-2.p-(1)=0

We add all the numbers together, and all the variables

(+p^2-3)-2.p-1=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

p^2-2p-4=0

a = 1; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·1·(-4)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*1}=\frac{2-2\sqrt{5}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*1}=\frac{2+2\sqrt{5}}{2}
$