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

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

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

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

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2 x^{2} + x = 1

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

Decimal Notation

x = 0.5, x = - 1

Solution steps

2 x^{2} + x = 1

Move

1

to the left side

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

- 9 - 1

(3 i) (- 2 i) (4 i)

\frac{3}{8} + \frac{4}{8}

3 x + 1 \leq 5

153 = \frac{P - 1900}{51.1503948}

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