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

The solutions to the equation x^2+2x=4x are x=2,x=0

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

The zeros of x^2+2x=4x are x=2,x=0

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x^{2} + 2 x = 4 x

x = 2, x = 0

Solution steps

x^{2} + 2 x = 4 x

Move

4 x

to the left side

x^{2} - 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

x_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

x = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

6 x^{2} + 2 x + 3 = 0

3 x^{2} + 5 = 41

x, x^{2} - x = 0

6 x^{2} + 2 x + 6 = 0

x^{2} - 12 x + 11 = 0

What are the solutions to the equation x^2+2x=4x ?
The solutions to the equation x^2+2x=4x are x=2,x=0
Find the zeros of x^2+2x=4x
The zeros of x^2+2x=4x are x=2,x=0