What are the solutions to the equation 2n^2+12n+18=0 ?

The solutions to the equation 2n^2+12n+18=0 are n=-3

Find the zeros of 2n^2+12n+18=0

The zeros of 2n^2+12n+18=0 are n=-3

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2 n^{2} + 12 n + 18 = 0

n = - 3

Solution steps

2 n^{2} + 12 n + 18 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 \cdot 18 = 0

n_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 2}

n = \frac{- 12}{2 \cdot 2}

\frac{- 12}{2 \cdot 2} = - 3

n = - 3

The solution to the quadratic equation is:

n = - 3

- 4 x^{2} + 2 = 0

x, 5 x^{2} + 6 x y + 5 y^{2} - 8 = 0

(x + 2)^{2} = 6

x, x^{2} + y^{2} + y + 3 x = - 10

x^{2} + \frac{x}{5} = 0

What are the solutions to the equation 2n^2+12n+18=0 ?
The solutions to the equation 2n^2+12n+18=0 are n=-3
Find the zeros of 2n^2+12n+18=0
The zeros of 2n^2+12n+18=0 are n=-3