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

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

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

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

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

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

Decimal Notation

x = 0.5, x = - 1

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- 1 - 3}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

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

8 x^{2} = 3 x + 6

y^{2} - 3 y + 15 = 0

x, \frac{1}{3} y - x^{2} + 2 x = - 5

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

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

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