(w+7)(w+5)=3

Solution for (w+7)(w+5)=3 equation:

(w+7)(w+5)=3

We move all terms to the left:

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

We multiply parentheses ..

(+w^2+5w+7w+35)-3=0

We get rid of parentheses

w^2+5w+7w+35-3=0

We add all the numbers together, and all the variables

w^2+12w+32=0

a = 1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*1}=\frac{-16}{2}
=-8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*1}=\frac{-8}{2}
=-4
$