What are the solutions to the equation 8n^2+4n-16=-n^2 ?

The solutions to the equation 8n^2+4n-16=-n^2 are n=(2(sqrt(37)-1))/9 ,n=-(2(1+sqrt(37)))/9

Find the zeros of 8n^2+4n-16=-n^2

The zeros of 8n^2+4n-16=-n^2 are n=(2(sqrt(37)-1))/9 ,n=-(2(1+sqrt(37)))/9

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8 n^{2} + 4 n - 16 = - n^{2}

n = \frac{2 ( 37 - 1 )}{9}, n = - \frac{2 ( 1 + 37 )}{9}

Decimal Notation

n = 1.12950 \dots, n = - 1.57394 \dots

Solution steps

8 n^{2} + 4 n - 16 = - n^{2}

Move

n^{2}

to the left side

9 n^{2} + 4 n - 16 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 9 ( - 16 )}{2 \cdot 9}

4^{2} - 4 \cdot 9 (- 16) = 437

n_{1, 2} = \frac{- 4 \pm 4 37}{2 \cdot 9}

Separate the solutions

n_{1} = \frac{- 4 + 4 37}{2 \cdot 9}, n_{2} = \frac{- 4 - 4 37}{2 \cdot 9}

n = \frac{- 4 + 4 37}{2 \cdot 9} : \frac{2 ( 37 - 1 )}{9}

n = \frac{- 4 - 4 37}{2 \cdot 9} : - \frac{2 ( 1 + 37 )}{9}

The solutions to the quadratic equation are:

n = \frac{2 ( 37 - 1 )}{9}, n = - \frac{2 ( 1 + 37 )}{9}

r^{2} - 2 r + 5 = 0

∣ 2 - 4 x ∣ = 6

∣ x - 5 ∣ < 7

2 (t + 3) = 4

12 x^{3} - 8 x^{2} - 20 x

What are the solutions to the equation 8n^2+4n-16=-n^2 ?
The solutions to the equation 8n^2+4n-16=-n^2 are n=(2(sqrt(37)-1))/9 ,n=-(2(1+sqrt(37)))/9
Find the zeros of 8n^2+4n-16=-n^2
The zeros of 8n^2+4n-16=-n^2 are n=(2(sqrt(37)-1))/9 ,n=-(2(1+sqrt(37)))/9