What are the solutions to the equation 8n^2+n-3=-3n ?

The solutions to the equation 8n^2+n-3=-3n are n=(-1+sqrt(7))/4 ,n=-(1+sqrt(7))/4

Find the zeros of 8n^2+n-3=-3n

The zeros of 8n^2+n-3=-3n are n=(-1+sqrt(7))/4 ,n=-(1+sqrt(7))/4

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8 n^{2} + n - 3 = - 3 n

n = \frac{- 1 + 7}{4}, n = - \frac{1 + 7}{4}

Decimal Notation

n = 0.41143 \dots, n = - 0.91143 \dots

Solution steps

8 n^{2} + n - 3 = - 3 n

Move

3 n

to the left side

8 n^{2} + 4 n - 3 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

4^{2} - 4 \cdot 8 (- 3) = 47

n_{1, 2} = \frac{- 4 \pm 4 7}{2 \cdot 8}

Separate the solutions

n_{1} = \frac{- 4 + 4 7}{2 \cdot 8}, n_{2} = \frac{- 4 - 4 7}{2 \cdot 8}

n = \frac{- 4 + 4 7}{2 \cdot 8} : \frac{- 1 + 7}{4}

n = \frac{- 4 - 4 7}{2 \cdot 8} : - \frac{1 + 7}{4}

The solutions to the quadratic equation are:

n = \frac{- 1 + 7}{4}, n = - \frac{1 + 7}{4}

6 (x - 3) + 5 (x + 4) = 15

10 - (\frac{( x - 15 )}{3})^{2} 9 < x \leq 24

- 2 a - 4 = - 8

a^{2} - 11 a + 18

lo g_{16} (64)

What are the solutions to the equation 8n^2+n-3=-3n ?
The solutions to the equation 8n^2+n-3=-3n are n=(-1+sqrt(7))/4 ,n=-(1+sqrt(7))/4
Find the zeros of 8n^2+n-3=-3n
The zeros of 8n^2+n-3=-3n are n=(-1+sqrt(7))/4 ,n=-(1+sqrt(7))/4