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

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

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

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

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

x = 2, x = 0

Solution steps

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

Move

3 x

to the left side

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

Move

x^{2}

to the left side

3 x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 6)^{2} - 4 \cdot 3 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 3}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 3}

x = \frac{- ( - 6 ) + 6}{2 \cdot 3} : 2

x = \frac{- ( - 6 ) - 6}{2 \cdot 3} : 0

The solutions to the quadratic equation are:

x = 2, x = 0

39.1179765 x - 4.9 x^{2} + 125 = 0

2 x^{2} - 1 x - 120 = 0

k, 2 k (k - 4) = 1

x^{2} + 3 x - 1000 = 0

x^{2} + 16 x + 128 = 0

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