What are the solutions to the equation 2x^2-4x-8=-x^2+x ?

The solutions to the equation 2x^2-4x-8=-x^2+x are x= 8/3 ,x=-1

Find the zeros of 2x^2-4x-8=-x^2+x

The zeros of 2x^2-4x-8=-x^2+x are x= 8/3 ,x=-1

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2 x^{2} - 4 x - 8 = - x^{2} + x

x = \frac{8}{3}, x = - 1

Decimal Notation

x = 2.66666 \dots, x = - 1

Solution steps

2 x^{2} - 4 x - 8 = - x^{2} + x

Move

x

to the left side

2 x^{2} - 5 x - 8 = - x^{2}

Move

x^{2}

to the left side

3 x^{2} - 5 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 8) = 11

x_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 3}, x_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 3}

x = \frac{- ( - 5 ) + 11}{2 \cdot 3} : \frac{8}{3}

x = \frac{- ( - 5 ) - 11}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = \frac{8}{3}, x = - 1

4 x^{2} + 9 = 73

2 x^{2} + 3 = - 7 x

x + x^{2} = 15

(2 x - 7) (x + 3) = 3

11 x^{2} + 22 x = 0

What are the solutions to the equation 2x^2-4x-8=-x^2+x ?
The solutions to the equation 2x^2-4x-8=-x^2+x are x= 8/3 ,x=-1
Find the zeros of 2x^2-4x-8=-x^2+x
The zeros of 2x^2-4x-8=-x^2+x are x= 8/3 ,x=-1