(20-2x)(40-2x)=629

Solution for (20-2x)(40-2x)=629 equation:

(20-2x)(40-2x)=629

We move all terms to the left:

(20-2x)(40-2x)-(629)=0

We add all the numbers together, and all the variables

(-2x+20)(-2x+40)-629=0

We multiply parentheses ..

(+4x^2-80x-40x+800)-629=0

We get rid of parentheses

4x^2-80x-40x+800-629=0

We add all the numbers together, and all the variables

4x^2-120x+171=0

a = 4; b = -120; c = +171;
Δ = b2-4ac
Δ = -1202-4·4·171
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-108}{2*4}=\frac{12}{8}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+108}{2*4}=\frac{228}{8}
=28+1/2
$