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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+w^2+3w-7w-21)-7=0

We get rid of parentheses

w^2+3w-7w-21-7=0

We add all the numbers together, and all the variables

w^2-4w-28=0

a = 1; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·1·(-28)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{2}}{2*1}=\frac{4-8\sqrt{2}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{2}}{2*1}=\frac{4+8\sqrt{2}}{2}
$