(8+x)(4+x)-(64)=0

Solution for (8+x)(4+x)-(64)=0 equation:

(8+x)(4+x)-(64)=0

We add all the numbers together, and all the variables

(x+8)(x+4)-64=0

We multiply parentheses ..

(+x^2+4x+8x+32)-64=0

We get rid of parentheses

x^2+4x+8x+32-64=0

We add all the numbers together, and all the variables

x^2+12x-32=0

a = 1; b = 12; c = -32;
Δ = b2-4ac
Δ = 122-4·1·(-32)
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{17}}{2*1}=\frac{-12-4\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{17}}{2*1}=\frac{-12+4\sqrt{17}}{2}
$