5(1-x)^2-6(x^2-3x-7)=x(x-3)-2x(x+5)-2

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Solution for 5(1-x)^2-6(x^2-3x-7)=x(x-3)-2x(x+5)-2 equation: 



5(1-x)^2-6(x^2-3x-7)=x(x-3)-2x(x+5)-2

We move all terms to the left:

5(1-x)^2-6(x^2-3x-7)-(x(x-3)-2x(x+5)-2)=0

We add all the numbers together, and all the variables

5(-1x+1)^2-6(x^2-3x-7)-(x(x-3)-2x(x+5)-2)=0

We multiply parentheses

-6x^2+5(-1x+1)^2+18x-(x(x-3)-2x(x+5)-2)+42=0



We calculate terms in parentheses: -(x(x-3)-2x(x+5)-2), so:

x(x-3)-2x(x+5)-2

We multiply parentheses

x^2-2x^2-3x-10x-2

We add all the numbers together, and all the variables

-1x^2-13x-2

Back to the equation:

-(-1x^2-13x-2)



We add all the numbers together, and all the variables

-6x^2-(-1x^2-13x-2)+18x+5(-1x+1)^2+42=0

We get rid of parentheses

-6x^2+1x^2+13x+18x+5(-1x+1)^2+2+42=0

We add all the numbers together, and all the variables

-5x^2+31x+5(-1x+1)^2+44=0

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

-5x^2+31x+5(-1x+1)^2=-44

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