1/4(40-8x)=19x+12-5x

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Solution for 1/4(40-8x)=19x+12-5x equation: 



1/4(40-8x)=19x+12-5x

We move all terms to the left:

1/4(40-8x)-(19x+12-5x)=0



Domain of the equation: 4(40-8x)!=0

x∈R



We add all the numbers together, and all the variables

1/4(-8x+40)-(14x+12)=0

We get rid of parentheses

1/4(-8x+40)-14x-12=0

We multiply all the terms by the denominator


-14x*4(-8x+40)-12*4(-8x+40)+1=0

Wy multiply elements

-56x^2(--48x(-+1=0

We use the square of the difference formula

-56x^2(+48x(-1=0

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