What are the solutions to the equation 2n^2+3n-5=0 ?

The solutions to the equation 2n^2+3n-5=0 are n=1,n=-5/2

Find the zeros of 2n^2+3n-5=0

The zeros of 2n^2+3n-5=0 are n=1,n=-5/2

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2 n^{2} + 3 n - 5 = 0

n = 1, n = - \frac{5}{2}

Decimal Notation

n = 1, n = - 2.5

Solution steps

2 n^{2} + 3 n - 5 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 5) = 7

n_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- 3 + 7}{2 \cdot 2}, n_{2} = \frac{- 3 - 7}{2 \cdot 2}

n = \frac{- 3 + 7}{2 \cdot 2} : 1

n = \frac{- 3 - 7}{2 \cdot 2} : - \frac{5}{2}

The solutions to the quadratic equation are:

n = 1, n = - \frac{5}{2}

24 y^{2} + 10 y - 2

x - 15 = 14

\frac{2 π}{\frac{2 π}{5}}

(x + 7)^{2} (x + 4) (x - 1) > 0

∣ 2 x - 4 + 5 x - 3 ∣ \geq 8

What are the solutions to the equation 2n^2+3n-5=0 ?
The solutions to the equation 2n^2+3n-5=0 are n=1,n=-5/2
Find the zeros of 2n^2+3n-5=0
The zeros of 2n^2+3n-5=0 are n=1,n=-5/2