What are the solutions to the equation 6+x-x^2=0 ?

The solutions to the equation 6+x-x^2=0 are x=-2,x=3

Find the zeros of 6+x-x^2=0

The zeros of 6+x-x^2=0 are x=-2,x=3

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6 + x - x^{2} = 0

x = - 2, x = 3

Solution steps

6 + x - x^{2} = 0

Write in the standard form

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

- x^{2} + x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 6}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 6 = 5

x_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 ( - 1 )}, x_{2} = \frac{- 1 - 5}{2 ( - 1 )}

x = \frac{- 1 + 5}{2 ( - 1 )} : - 2

x = \frac{- 1 - 5}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = - 2, x = 3

t^{2} - 9 t + 18 = 0

x^{2} + 7 x + 12 = 0

- 11 + 4 x^{2} = - 7 x - 2

5 x^{2} - 2 - x^{2} = 3 x^{2} - 3

6 x^{2} = 36 x - 150

What are the solutions to the equation 6+x-x^2=0 ?
The solutions to the equation 6+x-x^2=0 are x=-2,x=3
Find the zeros of 6+x-x^2=0
The zeros of 6+x-x^2=0 are x=-2,x=3