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

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

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

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

x = 4

Solution steps

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

Expand

(x - 2)^{2} + x (x - 3) : 2 x^{2} - 7 x + 4

Expand

3 (x + 4) (x - 3) - (x + 2) (x - 1) + 2 : 2 x^{2} + 2 x - 32

2 x^{2} - 7 x + 4 = 2 x^{2} + 2 x - 32

Move

4

to the right side

2 x^{2} - 7 x = 2 x^{2} + 2 x - 36

Subtract

2 x^{2} + 2 x

from both sides

2 x^{2} - 7 x - (2 x^{2} + 2 x) = 2 x^{2} + 2 x - 36 - (2 x^{2} + 2 x)

Simplify

- 9 x = - 36

Divide both sides by

- 9

x = 4

\frac{2 g}{5} = 12

8 - ∣ 2 x - 3 ∣ \geq 2

20 x^{3} y^{7} + 28 x^{4} y^{3} + 32 x^{4} y^{2}

8 - 9

(\frac{1}{a})^{3}

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