(16-2x)(10-2x)=80

Solution for (16-2x)(10-2x)=80 equation:

(16-2x)(10-2x)=80

We move all terms to the left:

(16-2x)(10-2x)-(80)=0

We add all the numbers together, and all the variables

(-2x+16)(-2x+10)-80=0

We multiply parentheses ..

(+4x^2-20x-32x+160)-80=0

We get rid of parentheses

4x^2-20x-32x+160-80=0

We add all the numbers together, and all the variables

4x^2-52x+80=0

a = 4; b = -52; c = +80;
Δ = b2-4ac
Δ = -522-4·4·80
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{89}}{2*4}=\frac{52-4\sqrt{89}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{89}}{2*4}=\frac{52+4\sqrt{89}}{8}
$