(p+3)(p-5)+7=0

Solution for (p+3)(p-5)+7=0 equation:

(p+3)(p-5)+7=0

We multiply parentheses ..

(+p^2-5p+3p-15)+7=0

We get rid of parentheses

p^2-5p+3p-15+7=0

We add all the numbers together, and all the variables

p^2-2p-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$