What are the solutions to the equation n^2=12-n ?

The solutions to the equation n^2=12-n are n=3,n=-4

Find the zeros of n^2=12-n

The zeros of n^2=12-n are n=3,n=-4

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n^{2} = 12 - n

n = 3, n = - 4

Solution steps

n^{2} = 12 - n

Move

n

to the left side

n^{2} + n = 12

Move

12

to the left side

n^{2} + n - 12 = 0

Solve with the quadratic formula

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

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

n_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 1 + 7}{2 \cdot 1}, n_{2} = \frac{- 1 - 7}{2 \cdot 1}

n = \frac{- 1 + 7}{2 \cdot 1} : 3

n = \frac{- 1 - 7}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

n = 3, n = - 4

x, 3 x^{2} = 1

5 x^{2} + 18 x = - 9

- 3 x^{2} - 31 x - 72 = - x

c^{2} = 5^{2} - 4^{2}

91 - q^{2} = 1 + q

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