What are the solutions to the equation 3x^2=(8+4x)(x-2) ?

The solutions to the equation 3x^2=(8+4x)(x-2) are x=4,x=-4

Find the zeros of 3x^2=(8+4x)(x-2)

The zeros of 3x^2=(8+4x)(x-2) are x=4,x=-4

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

x = 4, x = - 4

Solution steps

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

Expand

(8 + 4 x) (x - 2) : 4 x^{2} - 16

3 x^{2} = 4 x^{2} - 16

Move

4 x^{2}

to the left side

- x^{2} = - 16

Divide both sides by

- 1

x^{2} = 16

For

x^{2} = f (a)

the solutions are

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

x = 16, x = - 16

16 = 4

- 16 = - 4

x = 4, x = - 4

3 (4 x - 1)^{2} + 1 = - 16

- 0.5 x^{2} + 3.2 x - 2 = 0

3 x^{2} - 16 x

3 x^{2} - 15 x

r^{2} = 2 z - z^{2}

What are the solutions to the equation 3x^2=(8+4x)(x-2) ?
The solutions to the equation 3x^2=(8+4x)(x-2) are x=4,x=-4
Find the zeros of 3x^2=(8+4x)(x-2)
The zeros of 3x^2=(8+4x)(x-2) are x=4,x=-4