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

The solutions to the equation 2x^2-x=3x^2+x are x=-2,x=0

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

The zeros of 2x^2-x=3x^2+x are x=-2,x=0

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

x = - 2, x = 0

Solution steps

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

Move

x

to the left side

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

Move

3 x^{2}

to the left side

- x^{2} - 2 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = 0

(x^{2} - 3 x + 4) = 0

2 x^{2} + 18 x + 36 = 0

λ^{2} + 6 λ + 5 = 0

x^{2} - 7 = - 14

x^{2} - 7 = - 12

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