(2x-3)^2-(4x)^2/2x+8x/4=-9

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Solution for (2x-3)^2-(4x)^2/2x+8x/4=-9 equation: 



(2x-3)^2-(4x)^2/2x+8x/4=-9

We move all terms to the left:

(2x-3)^2-(4x)^2/2x+8x/4-(-9)=0



Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



determiningTheFunctionDomain
-4x^2/2x+(2x-3)^2+8x/4-(-9)=0

We add all the numbers together, and all the variables

-4x^2/2x+(2x-3)^2+8x/4+9=0

We calculate fractions


(-16x^2)/8x+16x^2/8x+(2x-3)^2+9=0

We multiply all the terms by the denominator


(-16x^2)+16x^2+((2x-3)^2)*8x+9*8x=0

We add all the numbers together, and all the variables

16x^2+(-16x^2)+((2x-3)^2)*8x+9*8x=0

Wy multiply elements

16x^2+(-16x^2)+((2x-3)^2)*8x+72x=0

We get rid of parentheses

16x^2-16x^2+((2x-3)^2)*8x+72x=0

We add all the numbers together, and all the variables

72x+((2x-3)^2)*8x=0

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