(x-10)(2x+4)=2x^2+8

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Solution for (x-10)(2x+4)=2x^2+8 equation: 



(x-10)(2x+4)=2x^2+8

We move all terms to the left:

(x-10)(2x+4)-(2x^2+8)=0

We get rid of parentheses

-2x^2+(x-10)(2x+4)-8=0

We multiply parentheses ..

-2x^2+(+2x^2+4x-20x-40)-8=0

We get rid of parentheses

-2x^2+2x^2+4x-20x-40-8=0

We add all the numbers together, and all the variables

-16x-48=0

We move all terms containing x to the left, all other terms to the right

-16x=48

x=48/-16

x=-3

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