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

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

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

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

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

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

Decimal Notation

n = 2.5, n = - 1

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} : \frac{5}{2}

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

The solutions to the quadratic equation are:

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

2 x - 7 > - 5

29 \cdot 3

4 x + 5 < 7

64 + 27 v^{3}

x + \frac{1}{2} = \frac{11}{3}

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