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

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

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

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

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

x = - 2, x = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = 1

16.5 \cdot \frac{1}{10}

8 x < 24 or x - 4 > 1

\frac{5 k ^{2} - 50 k + 45}{k ^{2} - 18 k + 81}

64 - 4 z^{2}

\frac{x}{12} = \frac{3}{4}

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