What are the solutions to the equation 2n^2+13n+18=-1 ?

The solutions to the equation 2n^2+13n+18=-1 are n=(-13+sqrt(17))/4 ,n=(-13-sqrt(17))/4

Find the zeros of 2n^2+13n+18=-1

The zeros of 2n^2+13n+18=-1 are n=(-13+sqrt(17))/4 ,n=(-13-sqrt(17))/4

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2 n^{2} + 13 n + 18 = - 1

n = \frac{- 13 + 17}{4}, n = \frac{- 13 - 17}{4}

Decimal Notation

n = - 2.21922 \dots, n = - 4.28077 \dots

Solution steps

2 n^{2} + 13 n + 18 = - 1

Move

1

to the left side

2 n^{2} + 13 n + 19 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 2 \cdot 19}{2 \cdot 2}

1 3^{2} - 4 \cdot 2 \cdot 19 = 17

n_{1, 2} = \frac{- 13 \pm 17}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 13 + 17}{2 \cdot 2}, n_{2} = \frac{- 13 - 17}{2 \cdot 2}

n = \frac{- 13 + 17}{2 \cdot 2} : \frac{- 13 + 17}{4}

n = \frac{- 13 - 17}{2 \cdot 2} : \frac{- 13 - 17}{4}

The solutions to the quadratic equation are:

n = \frac{- 13 + 17}{4}, n = \frac{- 13 - 17}{4}

5 x = - 5

16 k^{2} - 24 k + 9

\frac{1}{2} \cdot 10

\frac{7}{8} (x - 2) + \frac{5}{6} (x - 4) = 20 - \frac{7}{3} (x - 7)

72 9^{- \frac{1}{3}}

What are the solutions to the equation 2n^2+13n+18=-1 ?
The solutions to the equation 2n^2+13n+18=-1 are n=(-13+sqrt(17))/4 ,n=(-13-sqrt(17))/4
Find the zeros of 2n^2+13n+18=-1
The zeros of 2n^2+13n+18=-1 are n=(-13+sqrt(17))/4 ,n=(-13-sqrt(17))/4