What are the solutions to the equation n^2-n-5=-2n^2-2n^2 ?

The solutions to the equation n^2-n-5=-2n^2-2n^2 are n=(1+sqrt(101))/(10),n=(1-sqrt(101))/(10)

Find the zeros of n^2-n-5=-2n^2-2n^2

The zeros of n^2-n-5=-2n^2-2n^2 are n=(1+sqrt(101))/(10),n=(1-sqrt(101))/(10)

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n^{2} - n - 5 = - 2 n^{2} - 2 n^{2}

n = \frac{1 + 101}{10}, n = \frac{1 - 101}{10}

Decimal Notation

n = 1.10498 \dots, n = - 0.90498 \dots

Solution steps

n^{2} - n - 5 = - 2 n^{2} - 2 n^{2}

Refine

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

Move

4 n^{2}

to the left side

5 n^{2} - n - 5 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 1)^{2} - 4 \cdot 5 (- 5) = 101

n_{1, 2} = \frac{- ( - 1 ) \pm 101}{2 \cdot 5}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 101}{2 \cdot 5}, n_{2} = \frac{- ( - 1 ) - 101}{2 \cdot 5}

n = \frac{- ( - 1 ) + 101}{2 \cdot 5} : \frac{1 + 101}{10}

n = \frac{- ( - 1 ) - 101}{2 \cdot 5} : \frac{1 - 101}{10}

The solutions to the quadratic equation are:

n = \frac{1 + 101}{10}, n = \frac{1 - 101}{10}

(0.036 - x) (7.1 \times 1 0^{- 4}) = x^{2}

2 x^{2} + 16 x + 26 = 0

6 0^{2} + 8 0^{2} = x^{2}

x^{2} + 24 x + 150 = 0

49 x^{2} - 16 = 20

What are the solutions to the equation n^2-n-5=-2n^2-2n^2 ?
The solutions to the equation n^2-n-5=-2n^2-2n^2 are n=(1+sqrt(101))/(10),n=(1-sqrt(101))/(10)
Find the zeros of n^2-n-5=-2n^2-2n^2
The zeros of n^2-n-5=-2n^2-2n^2 are n=(1+sqrt(101))/(10),n=(1-sqrt(101))/(10)