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

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

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

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

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2 - x^{2} - x = 0

x = - 2, x = 1

Solution steps

2 - x^{2} - x = 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 )} : - 2

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

The solutions to the quadratic equation are:

x = - 2, x = 1

3 (5 x + 1) = 17 x - 1

- 5 x = - \frac{3}{2} + \frac{1}{4} (- 4 x - 8)

(\frac{3}{7} + \frac{2}{7}) \cdot \frac{1}{2}

x^{2} y^{2} - 36

∣ 7 x - 56 ∣ = 0

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