What are the solutions to the equation n^2-3n+8n=4 ?

The solutions to the equation n^2-3n+8n=4 are n=(-5+sqrt(41))/2 ,n=(-5-sqrt(41))/2

Find the zeros of n^2-3n+8n=4

The zeros of n^2-3n+8n=4 are n=(-5+sqrt(41))/2 ,n=(-5-sqrt(41))/2

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

n = \frac{- 5 + 41}{2}, n = \frac{- 5 - 41}{2}

Decimal Notation

n = 0.70156 \dots, n = - 5.70156 \dots

Solution steps

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

Refine

n^{2} + 5 n = 4

Move

4

to the left side

n^{2} + 5 n - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 4) = 41

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = \frac{- 5 + 41}{2}, n = \frac{- 5 - 41}{2}

100 x^{2} - 60 x + 9 = 0

(a + 4) (a - 2) = 55

2 x (x - 6) = 10

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

x^{2} + 24 x - 144 = 0

What are the solutions to the equation n^2-3n+8n=4 ?
The solutions to the equation n^2-3n+8n=4 are n=(-5+sqrt(41))/2 ,n=(-5-sqrt(41))/2
Find the zeros of n^2-3n+8n=4
The zeros of n^2-3n+8n=4 are n=(-5+sqrt(41))/2 ,n=(-5-sqrt(41))/2