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,x= 1/2

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

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

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

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

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 )} : - 1

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

The solutions to the quadratic equation are:

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

3 ∣ 2 x - 1 ∣ + 2 = 5

x, h (x) = f (g)

2 a^{2} - 3 ab - 5 b^{2}

- \frac{5}{4} - 2

x^{2} + 8 x + 12 \geq 0

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