What are the solutions to the equation n^2+13n+22=7 ?

The solutions to the equation n^2+13n+22=7 are n=(-13+sqrt(109))/2 ,n=(-13-sqrt(109))/2

Find the zeros of n^2+13n+22=7

The zeros of n^2+13n+22=7 are n=(-13+sqrt(109))/2 ,n=(-13-sqrt(109))/2

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n^{2} + 13 n + 22 = 7

n = \frac{- 13 + 109}{2}, n = \frac{- 13 - 109}{2}

Decimal Notation

n = - 1.27984 \dots, n = - 11.72015 \dots

Solution steps

n^{2} + 13 n + 22 = 7

Move

7

to the left side

n^{2} + 13 n + 15 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 15}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 15 = 109

n_{1, 2} = \frac{- 13 \pm 109}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 13 + 109}{2 \cdot 1}, n_{2} = \frac{- 13 - 109}{2 \cdot 1}

n = \frac{- 13 + 109}{2 \cdot 1} : \frac{- 13 + 109}{2}

n = \frac{- 13 - 109}{2 \cdot 1} : \frac{- 13 - 109}{2}

The solutions to the quadratic equation are:

n = \frac{- 13 + 109}{2}, n = \frac{- 13 - 109}{2}

(\frac{( x + 2 )}{2}) + (\frac{( x - 3 )}{5}) = (\frac{6}{10}) + 12

5 x^{2} + 50 x y + 125 y^{2}

6 x - 2 \geq 9 x - 14

(6 x^{5})^{2}

∣ 6 - 4 x ∣ = 14

What are the solutions to the equation n^2+13n+22=7 ?
The solutions to the equation n^2+13n+22=7 are n=(-13+sqrt(109))/2 ,n=(-13-sqrt(109))/2
Find the zeros of n^2+13n+22=7
The zeros of n^2+13n+22=7 are n=(-13+sqrt(109))/2 ,n=(-13-sqrt(109))/2