(20-x)(8+x)=192

Solution for (20-x)(8+x)=192 equation:

(20-x)(8+x)=192

We move all terms to the left:

(20-x)(8+x)-(192)=0

We add all the numbers together, and all the variables

(-1x+20)(x+8)-192=0

We multiply parentheses ..

(-1x^2-8x+20x+160)-192=0

We get rid of parentheses

-1x^2-8x+20x+160-192=0

We add all the numbers together, and all the variables

-1x^2+12x-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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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