What are the solutions to the equation 5n^2+2n-8=0 ?

The solutions to the equation 5n^2+2n-8=0 are n=(-1+sqrt(41))/5 ,n=-(1+sqrt(41))/5

Find the zeros of 5n^2+2n-8=0

The zeros of 5n^2+2n-8=0 are n=(-1+sqrt(41))/5 ,n=-(1+sqrt(41))/5

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5 n^{2} + 2 n - 8 = 0

n = \frac{- 1 + 41}{5}, n = - \frac{1 + 41}{5}

Decimal

n = 1.08062 \dots, n = - 1.48062 \dots

Solution steps

5 n^{2} + 2 n - 8 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 8 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 8) = 241

n_{1, 2} = \frac{- 2 \pm 2 41}{2 \cdot 5}

Separate the solutions

n_{1} = \frac{- 2 + 2 41}{2 \cdot 5}, n_{2} = \frac{- 2 - 2 41}{2 \cdot 5}

n = \frac{- 2 + 2 41}{2 \cdot 5} : \frac{- 1 + 41}{5}

n = \frac{- 2 - 2 41}{2 \cdot 5} : - \frac{1 + 41}{5}

The solutions to the quadratic equation are:

n = \frac{- 1 + 41}{5}, n = - \frac{1 + 41}{5}

(x^{2} - 3 x + 9) = 0

x, d = \frac{1}{3} x^{2} t

63.97 d^{2} + 20.626 d - 37.02 = 0

x^{2} + (x - 2)^{2} = 164

5.3 \cdot 1 0^{- 12} = (1 \cdot 1 0^{- 4} + s) (s)

What are the solutions to the equation 5n^2+2n-8=0 ?
The solutions to the equation 5n^2+2n-8=0 are n=(-1+sqrt(41))/5 ,n=-(1+sqrt(41))/5
Find the zeros of 5n^2+2n-8=0
The zeros of 5n^2+2n-8=0 are n=(-1+sqrt(41))/5 ,n=-(1+sqrt(41))/5