What are the solutions to the equation 2n^2+5n-9=0 ?

The solutions to the equation 2n^2+5n-9=0 are n=(-5+sqrt(97))/4 ,n=(-5-sqrt(97))/4

Find the zeros of 2n^2+5n-9=0

The zeros of 2n^2+5n-9=0 are n=(-5+sqrt(97))/4 ,n=(-5-sqrt(97))/4

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2 n^{2} + 5 n - 9 = 0

n = \frac{- 5 + 97}{4}, n = \frac{- 5 - 97}{4}

Decimal Notation

n = 1.21221 \dots, n = - 3.71221 \dots

Solution steps

2 n^{2} + 5 n - 9 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 9) = 97

n_{1, 2} = \frac{- 5 \pm 97}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 5 + 97}{2 \cdot 2}, n_{2} = \frac{- 5 - 97}{2 \cdot 2}

n = \frac{- 5 + 97}{2 \cdot 2} : \frac{- 5 + 97}{4}

n = \frac{- 5 - 97}{2 \cdot 2} : \frac{- 5 - 97}{4}

The solutions to the quadratic equation are:

n = \frac{- 5 + 97}{4}, n = \frac{- 5 - 97}{4}

x^{2} + \frac{5}{12} x - \frac{1}{6} = 0

(x + 5) (x - 6) = - 26

x, 5 x^{2} - 6 x y + 5 y^{2} = 4

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

x = 9 - x^{2}

What are the solutions to the equation 2n^2+5n-9=0 ?
The solutions to the equation 2n^2+5n-9=0 are n=(-5+sqrt(97))/4 ,n=(-5-sqrt(97))/4
Find the zeros of 2n^2+5n-9=0
The zeros of 2n^2+5n-9=0 are n=(-5+sqrt(97))/4 ,n=(-5-sqrt(97))/4