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

l + 15.2 = - 4.5

\frac{x + 3}{4} - \frac{x - 4}{9} = \frac{1}{2} - \frac{x + 1}{4} + \frac{2 x + 1}{9}

x, 2 x - 3 y - 12 = 0

\frac{5 r + 7}{3} = 3 - \frac{r}{6}

6 = \frac{x}{4} + 2

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