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