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

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

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

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

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

x = 3, x = - 3

Decimal Notation

x = 1.73205 \dots, x = - 1.73205 \dots

Solution steps

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

Expand

- x^{2} + 2 (x + 3) : - x^{2} + 2 x + 6

- x^{2} + 2 x + 6 = 3 + 2 x

Move

6

to the right side

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

Move

2 x

to the left side

- x^{2} = - 3

Divide both sides by

- 1

x^{2} = 3

For

x^{2} = f (a)

the solutions are

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

x = 3, x = - 3

p^{2} + 13 p - 30 = 0

x^{2} + 4 a x - 12 a^{2} = 0

7^{2} + 2^{2} = c^{2}

x, y = - 0.00002 x^{2} + 0.0136 x + 0.1399

x, x^{2} + y^{2} = z^{2} (1 - z)

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