(w+2)(w+4)=63

Solution for (w+2)(w+4)=63 equation:

(w+2)(w+4)=63

We move all terms to the left:

(w+2)(w+4)-(63)=0

We multiply parentheses ..

(+w^2+4w+2w+8)-63=0

We get rid of parentheses

w^2+4w+2w+8-63=0

We add all the numbers together, and all the variables

w^2+6w-55=0

a = 1; b = 6; c = -55;
Δ = b2-4ac
Δ = 62-4·1·(-55)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-16}{2*1}=\frac{-22}{2}
=-11
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+16}{2*1}=\frac{10}{2}
=5
$