What are the solutions to the equation n(2n-3)=2 ?

The solutions to the equation n(2n-3)=2 are n=2,n=-1/2

Find the zeros of n(2n-3)=2

The zeros of n(2n-3)=2 are n=2,n=-1/2

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n (2 n - 3) = 2

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

Decimal Notation

n = 2, n = - 0.5

Solution steps

n (2 n - 3) = 2

Expand

n (2 n - 3) : 2 n^{2} - 3 n

2 n^{2} - 3 n = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 2}, n_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 2}

n = \frac{- ( - 3 ) + 5}{2 \cdot 2} : 2

n = \frac{- ( - 3 ) - 5}{2 \cdot 2} : - \frac{1}{2}

The solutions to the quadratic equation are:

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

\frac{( 1 + 3 i )}{( 2 + i )}

a - 3 = - 4 a + 24

36 - 12 x + x^{2}

x^{2} - 11 x = 6

2 5^{\frac{1}{2}}

What are the solutions to the equation n(2n-3)=2 ?
The solutions to the equation n(2n-3)=2 are n=2,n=-1/2
Find the zeros of n(2n-3)=2
The zeros of n(2n-3)=2 are n=2,n=-1/2