4(x+5)-0,5(x-3)(6-2x)=x2+1

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Solution for 4(x+5)-0,5(x-3)(6-2x)=x2+1 equation: 



4(x+5)-0.5(x-3)(6-2x)=x2+1

We move all terms to the left:

4(x+5)-0.5(x-3)(6-2x)-(x2+1)=0

We add all the numbers together, and all the variables

-(+x^2+1)+4(x+5)-0.5(x-3)(-2x+6)=0

We multiply parentheses

-(+x^2+1)+4x-0.5(x-3)(-2x+6)+20=0

We get rid of parentheses

-x^2+4x-0.5(x-3)(-2x+6)-1+20=0

We multiply parentheses ..

-x^2-0.5(-2x^2+6x+6x-18)+4x-1+20=0

We add all the numbers together, and all the variables

-1x^2-0.5(-2x^2+6x+6x-18)+4x+19=0

We multiply parentheses

-1x^2+1x^2-3x-3x+4x+9+19=0

We add all the numbers together, and all the variables

-2x+28=0

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

-2x=-28

x=-28/-2

x=+14

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