What is expand x^2-(3x+(x(x+1)+4(x^2-1)-x^2)=0) ?

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

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

expand

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

x^{2} - 3 x - = 0

Solution steps

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

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

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

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

Expand

- = 0 \cdot (x (x + 1) + 4 (x^{2} - 1) - x^{2}) : - = 0

x^{2} - 3 x - = 0

(x - 1 + 6 i) (x - 1 - 6 i) (x - 1) (x - 2) (2 x - 1)

\frac{s ^{2} + s + 2}{( s + 1 ) ^{3}}

x (θ - sin (θ))

(x - 14) (x - \frac{5}{2}) (x - \frac{9}{2})

\frac{( - z + 6 ) ^{3}}{54}

What is expand x^2-(3x+(x(x+1)+4(x^2-1)-x^2)=0) ?
The solution to expand x^2-(3x+(x(x+1)+4(x^2-1)-x^2)=0) is x^2-3x-=0