What are the solutions to the equation -2x^2+2x+4=0 ?

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

Find the zeros of -2x^2+2x+4=0

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

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- 2 x^{2} + 2 x + 4 = 0

x = - 1, x = 2

Solution steps

- 2 x^{2} + 2 x + 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 2 + 6}{2 ( - 2 )} : - 1

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

The solutions to the quadratic equation are:

x = - 1, x = 2

\frac{7}{8} \cdot \frac{2}{3}

2 x^{2} = 5 x - 2

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

2 x = 24

2 x = 12

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