3-3x-2(1-x)=7x(x-4)+29

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Solution for 3-3x-2(1-x)=7x(x-4)+29 equation: 



3-3x-2(1-x)=7x(x-4)+29

We move all terms to the left:

3-3x-2(1-x)-(7x(x-4)+29)=0

We add all the numbers together, and all the variables

-3x-2(-1x+1)-(7x(x-4)+29)+3=0

We multiply parentheses

-3x+2x-(7x(x-4)+29)-2+3=0



We calculate terms in parentheses: -(7x(x-4)+29), so:

7x(x-4)+29

We multiply parentheses

7x^2-28x+29

Back to the equation:

-(7x^2-28x+29)



We add all the numbers together, and all the variables

-1x-(7x^2-28x+29)+1=0

We get rid of parentheses

-7x^2-1x+28x-29+1=0

We add all the numbers together, and all the variables

-7x^2+27x-28=0

a = -7; b = 27; c = -28;
Δ = b2-4ac
Δ = 272-4·(-7)·(-28)
Δ = -55
Delta is less than zero, so there is no solution for the equation

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