What are the solutions to the equation x-x^2+12=0 ?

The solutions to the equation x-x^2+12=0 are x=-3,x=4

Find the zeros of x-x^2+12=0

The zeros of x-x^2+12=0 are x=-3,x=4

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x - x^{2} + 12 = 0

x = - 3, x = 4

Solution steps

x - x^{2} + 12 = 0

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 12}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 12 = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 ( - 1 )}, x_{2} = \frac{- 1 - 7}{2 ( - 1 )}

x = \frac{- 1 + 7}{2 ( - 1 )} : - 3

x = \frac{- 1 - 7}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

x = - 3, x = 4

123

1.4 \cdot 1 0^{5}

19.6 e^{0.0138 (0)}

x^{2} - 11 x + 31 = 0

3 + 4 + 6 i

What are the solutions to the equation x-x^2+12=0 ?
The solutions to the equation x-x^2+12=0 are x=-3,x=4
Find the zeros of x-x^2+12=0
The zeros of x-x^2+12=0 are x=-3,x=4