What are the solutions to the equation -2x^2+x+1=0 ?

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

Find the zeros of -2x^2+x+1=0

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

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

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

Decimal Notation

x = - 0.5, x = 1

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

Separate the solutions

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

x = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

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

The solutions to the quadratic equation are:

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

8 (x - 1) = - 2 (x - 41)

5 v^{2} - 16 = 16 v

3 x^{2} + 6 x = 45

\frac{2}{3} \cdot \frac{2}{3}

lo g_{4} (16) = 4

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