What are the solutions to the equation x^2-3x=3x^2-4x ?

The solutions to the equation x^2-3x=3x^2-4x are x=0,x= 1/2

Find the zeros of x^2-3x=3x^2-4x

The zeros of x^2-3x=3x^2-4x are x=0,x= 1/2

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

x = 0, x = \frac{1}{2}

Decimal Notation

x = 0, x = 0.5

Solution steps

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

Move

4 x

to the left side

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

Move

3 x^{2}

to the left side

- 2 x^{2} + x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 0, x = \frac{1}{2}

3 n^{2} - 29 n + 855 = 0

x^{2} + 9 = ∣ x ∣ + 3

e^{- x^{2}} \cdot e^{x^{2}}

5 x^{2} + 29 x + 20

3 (x + 1) (x - 1) + 7 x = 5 - 3 x

What are the solutions to the equation x^2-3x=3x^2-4x ?
The solutions to the equation x^2-3x=3x^2-4x are x=0,x= 1/2
Find the zeros of x^2-3x=3x^2-4x
The zeros of x^2-3x=3x^2-4x are x=0,x= 1/2