What are the solutions to the equation x^2-4x+4+x^2+4x+4=3x^2-x+x-4 ?

The solutions to the equation x^2-4x+4+x^2+4x+4=3x^2-x+x-4 are x=2sqrt(3),x=-2sqrt(3)

Find the zeros of x^2-4x+4+x^2+4x+4=3x^2-x+x-4

The zeros of x^2-4x+4+x^2+4x+4=3x^2-x+x-4 are x=2sqrt(3),x=-2sqrt(3)

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x^{2} - 4 x + 4 + x^{2} + 4 x + 4 = 3 x^{2} - x + x - 4

x = 23, x = - 23

Decimal Notation

x = 3.46410 \dots, x = - 3.46410 \dots

Solution steps

x^{2} - 4 x + 4 + x^{2} + 4 x + 4 = 3 x^{2} - x + x - 4

Move

4

to the right side

x^{2} - 4 x + x^{2} + 4 x + 4 = 3 x^{2} - 8

Move

4

to the right side

x^{2} - 4 x + x^{2} + 4 x = 3 x^{2} - 12

Move

3 x^{2}

to the left side

- x^{2} = - 12

Divide both sides by

- 1

x^{2} = 12

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 12, x = - 12

12 = 23

- 12 = - 23

x = 23, x = - 23

7 x^{2} - 3 x + 10 = 0

9 x^{2} - 42 x + 45 = 0

x^{2} + x - 48 = 0

z^{2} + 5 z - 3 = 0

What are the solutions to the equation x^2-4x+4+x^2+4x+4=3x^2-x+x-4 ?
The solutions to the equation x^2-4x+4+x^2+4x+4=3x^2-x+x-4 are x=2sqrt(3),x=-2sqrt(3)
Find the zeros of x^2-4x+4+x^2+4x+4=3x^2-x+x-4
The zeros of x^2-4x+4+x^2+4x+4=3x^2-x+x-4 are x=2sqrt(3),x=-2sqrt(3)