2(x)(10-2x)+2(x)(16-2x)+(10-2x)(16-2x)=144

Solution for 2(x)(10-2x)+2(x)(16-2x)+(10-2x)(16-2x)=144 equation:

2(x)(10-2x)+2(x)(16-2x)+(10-2x)(16-2x)=144

We move all terms to the left:

2(x)(10-2x)+2(x)(16-2x)+(10-2x)(16-2x)-(144)=0

We add all the numbers together, and all the variables

2x(-2x+10)+2x(-2x+16)+(-2x+10)(-2x+16)-144=0

We multiply parentheses

-4x^2-4x^2+20x+32x+(-2x+10)(-2x+16)-144=0

We multiply parentheses ..

-4x^2-4x^2+(+4x^2-32x-20x+160)+20x+32x-144=0

We add all the numbers together, and all the variables

-8x^2+(+4x^2-32x-20x+160)+52x-144=0

We get rid of parentheses

-8x^2+4x^2-32x-20x+52x+160-144=0

We add all the numbers together, and all the variables

-4x^2+16=0

a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-4}=\frac{-16}{-8}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8}
=-2
$