What are the solutions to the equation 1=n-1n-4+1n^2-n-12 ?

The solutions to the equation 1=n-1n-4+1n^2-n-12 are n=(1+sqrt(69))/2 ,n=(1-sqrt(69))/2

Find the zeros of 1=n-1n-4+1n^2-n-12

The zeros of 1=n-1n-4+1n^2-n-12 are n=(1+sqrt(69))/2 ,n=(1-sqrt(69))/2

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1 = n - 1 n - 4 + 1 n^{2} - n - 12

n = \frac{1 + 69}{2}, n = \frac{1 - 69}{2}

Decimal Notation

n = 4.65331 \dots, n = - 3.65331 \dots

Solution steps

1 = n - 1 \cdot n - 4 + 1 \cdot n^{2} - n - 12

Refine

1 = n^{2} - n - 16

Switch sides

n^{2} - n - 16 = 1

Move

1

to the left side

n^{2} - n - 17 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 17 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 17) = 69

n_{1, 2} = \frac{- ( - 1 ) \pm 69}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 69}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 69}{2 \cdot 1}

n = \frac{- ( - 1 ) + 69}{2 \cdot 1} : \frac{1 + 69}{2}

n = \frac{- ( - 1 ) - 69}{2 \cdot 1} : \frac{1 - 69}{2}

The solutions to the quadratic equation are:

n = \frac{1 + 69}{2}, n = \frac{1 - 69}{2}

- x^{2} = - 10

6 x^{2} - 42 x = 0

0 = 168 π x - 6 π^{2} x^{2}

x, x^{2} + x y = 0

- x^{2} + x + 1 = 0

What are the solutions to the equation 1=n-1n-4+1n^2-n-12 ?
The solutions to the equation 1=n-1n-4+1n^2-n-12 are n=(1+sqrt(69))/2 ,n=(1-sqrt(69))/2
Find the zeros of 1=n-1n-4+1n^2-n-12
The zeros of 1=n-1n-4+1n^2-n-12 are n=(1+sqrt(69))/2 ,n=(1-sqrt(69))/2