(p-2)(p-5)=3

Solution for (p-2)(p-5)=3 equation:

(p-2)(p-5)=3

We move all terms to the left:

(p-2)(p-5)-(3)=0

We multiply parentheses ..

(+p^2-5p-2p+10)-3=0

We get rid of parentheses

p^2-5p-2p+10-3=0

We add all the numbers together, and all the variables

p^2-7p+7=0

a = 1; b = -7; c = +7;
Δ = b2-4ac
Δ = -72-4·1·7
Δ = 21
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{21}}{2*1}=\frac{7-\sqrt{21}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{21}}{2*1}=\frac{7+\sqrt{21}}{2}
$