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

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

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

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

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2 + x - x^{2} = 0

x = - 1, x = 2

Solution steps

2 + x - x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 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 )} : - 1

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

The solutions to the quadratic equation are:

x = - 1, x = 2

a^{2} - a + 2 = 0

b^{2} = 64

24 x - 6 x^{2} = 0

(n - 7)^{2} = 2 n

x^{2} + 2 x - 5 = 0

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