What is the answer to x^2-(3x+(x(x+1)+4(x^2-1)-4x^2))=0 ?

The answer to x^2-(3x+(x(x+1)+4(x^2-1)-4x^2))=0 is x=1

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

x^{2} - (3 x + (x (x + 1) + 4 (x^{2} - 1) - 4 x^{2})) = 0

x = 1

Solution steps

x^{2} - (3 x + (x (x + 1) + 4 (x^{2} - 1) - 4 x^{2})) = 0

x^{2} - (3 x + x (x + 1) + 4 (x^{2} - 1) - 4 x^{2}) = 0

Expand

x^{2} - (3 x + x (x + 1) + 4 (x^{2} - 1) - 4 x^{2}) : - 4 x + 4

- 4 x + 4 = 0

Move

4

to the right side

- 4 x = - 4

Divide both sides by

- 4

x = 1

\frac{3 x + 18}{7} = 27

x, 4 x + y = 17

x, 4 x + y = - 7

x, 3 x - \frac{1}{2} y = 0

x + 3 - x = - 10

What is the answer to x^2-(3x+(x(x+1)+4(x^2-1)-4x^2))=0 ?
The answer to x^2-(3x+(x(x+1)+4(x^2-1)-4x^2))=0 is x=1
Solve for x: x^2-(3x+(x(x+1)+4(x^2-1)-4x^2))=0
The solution is x=1