What are the solutions to the equation n^2+17n+22=7 ?

The solutions to the equation n^2+17n+22=7 are n=(-17+sqrt(229))/2 ,n=(-17-sqrt(229))/2

Find the zeros of n^2+17n+22=7

The zeros of n^2+17n+22=7 are n=(-17+sqrt(229))/2 ,n=(-17-sqrt(229))/2

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n^{2} + 17 n + 22 = 7

n = \frac{- 17 + 229}{2}, n = \frac{- 17 - 229}{2}

Decimal Notation

n = - 0.93362 \dots, n = - 16.06637 \dots

Solution steps

n^{2} + 17 n + 22 = 7

Move

7

to the left side

n^{2} + 17 n + 15 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

1 7^{2} - 4 \cdot 1 \cdot 15 = 229

n_{1, 2} = \frac{- 17 \pm 229}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 17 + 229}{2 \cdot 1}, n_{2} = \frac{- 17 - 229}{2 \cdot 1}

n = \frac{- 17 + 229}{2 \cdot 1} : \frac{- 17 + 229}{2}

n = \frac{- 17 - 229}{2 \cdot 1} : \frac{- 17 - 229}{2}

The solutions to the quadratic equation are:

n = \frac{- 17 + 229}{2}, n = \frac{- 17 - 229}{2}

a, c^{2} = a^{2} + b^{2}

4 + 3 x^{2} = 7

- m^{2} + 3 m - 2 = 0

2 x^{2} + (4 b - a) x - 2 ab = 0

x^{2} - 5 x + 6

What are the solutions to the equation n^2+17n+22=7 ?
The solutions to the equation n^2+17n+22=7 are n=(-17+sqrt(229))/2 ,n=(-17-sqrt(229))/2
Find the zeros of n^2+17n+22=7
The zeros of n^2+17n+22=7 are n=(-17+sqrt(229))/2 ,n=(-17-sqrt(229))/2