What are the solutions to the equation 3x^2-12=(x-4)^2+8x ?

The solutions to the equation 3x^2-12=(x-4)^2+8x are x=sqrt(14),x=-sqrt(14)

Find the zeros of 3x^2-12=(x-4)^2+8x

The zeros of 3x^2-12=(x-4)^2+8x are x=sqrt(14),x=-sqrt(14)

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3 x^{2} - 12 = (x - 4)^{2} + 8 x

x = 14, x = - 14

Decimal Notation

x = 3.74165 \dots, x = - 3.74165 \dots

Solution steps

3 x^{2} - 12 = (x - 4)^{2} + 8 x

Expand

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

3 x^{2} - 12 = x^{2} + 16

Move

12

to the right side

3 x^{2} = x^{2} + 28

Move

x^{2}

to the left side

2 x^{2} = 28

Divide both sides by

2

x^{2} = 14

For

x^{2} = f (a)

the solutions are

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

x = 14, x = - 14

4 r (r - 3) - 3 (r - 3) = 0

x^{2} - 18 x - 32 = 0

k^{2} + 8 k + 12 = 0

- 6 x + 5 x^{2} = - 8

x, 5 x^{2} + 5 y^{2} = 0

What are the solutions to the equation 3x^2-12=(x-4)^2+8x ?
The solutions to the equation 3x^2-12=(x-4)^2+8x are x=sqrt(14),x=-sqrt(14)
Find the zeros of 3x^2-12=(x-4)^2+8x
The zeros of 3x^2-12=(x-4)^2+8x are x=sqrt(14),x=-sqrt(14)