What are the solutions to the equation 2n^2-n-4=2 ?

The solutions to the equation 2n^2-n-4=2 are n=2,n=-3/2

Find the zeros of 2n^2-n-4=2

The zeros of 2n^2-n-4=2 are n=2,n=-3/2

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2 n^{2} - n - 4 = 2

n = 2, n = - \frac{3}{2}

Decimal Notation

n = 2, n = - 1.5

Solution steps

2 n^{2} - n - 4 = 2

Move

2

to the left side

2 n^{2} - n - 6 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 6) = 7

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 2, n = - \frac{3}{2}

\frac{3}{7} = \frac{9}{28}

1.49 m \leq 8.94

lo g_{3} (x) = 4

3 x^{\frac{3}{2}} - 9 x^{\frac{1}{2}} - 6 x^{- \frac{1}{2}}

∣ x + 4 ∣ > 1

What are the solutions to the equation 2n^2-n-4=2 ?
The solutions to the equation 2n^2-n-4=2 are n=2,n=-3/2
Find the zeros of 2n^2-n-4=2
The zeros of 2n^2-n-4=2 are n=2,n=-3/2