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

The solutions to the equation x^2+2x-4=2x are x=2,x=-2

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

The zeros of x^2+2x-4=2x are x=2,x=-2

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

x = 2, x = - 2

Solution steps

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

Move

4

to the right side

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

Move

2 x

to the left side

x^{2} = 4

For

x^{2} = f (a)

the solutions are

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

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

20 \cdot \frac{4}{5}

- \frac{1}{3} - \frac{( - \frac{1}{2} )}{( - \frac{1}{3} )}

(1 + i)^{2} (1 - i)^{3}

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

4 x^{2} + 20 x y + 25 y^{2}

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