What are the solutions to the equation 2n^2-n-7=0 ?

The solutions to the equation 2n^2-n-7=0 are n=(1+sqrt(57))/4 ,n=(1-sqrt(57))/4

Find the zeros of 2n^2-n-7=0

The zeros of 2n^2-n-7=0 are n=(1+sqrt(57))/4 ,n=(1-sqrt(57))/4

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2 n^{2} - n - 7 = 0

n = \frac{1 + 57}{4}, n = \frac{1 - 57}{4}

Decimal Notation

n = 2.13745 \dots, n = - 1.63745 \dots

Solution steps

2 n^{2} - n - 7 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 7) = 57

n_{1, 2} = \frac{- ( - 1 ) \pm 57}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 57}{2 \cdot 2}, n_{2} = \frac{- ( - 1 ) - 57}{2 \cdot 2}

n = \frac{- ( - 1 ) + 57}{2 \cdot 2} : \frac{1 + 57}{4}

n = \frac{- ( - 1 ) - 57}{2 \cdot 2} : \frac{1 - 57}{4}

The solutions to the quadratic equation are:

n = \frac{1 + 57}{4}, n = \frac{1 - 57}{4}

18 x^{2} - 1 = 0

82 \cdot 82

x^{2} - 121

x^{2} \leq - 3 x - 2

\frac{2 \cdot ( x - 12 )}{3} - 4 \cdot (x - 2) = \frac{7 - x}{3} + 8

What are the solutions to the equation 2n^2-n-7=0 ?
The solutions to the equation 2n^2-n-7=0 are n=(1+sqrt(57))/4 ,n=(1-sqrt(57))/4
Find the zeros of 2n^2-n-7=0
The zeros of 2n^2-n-7=0 are n=(1+sqrt(57))/4 ,n=(1-sqrt(57))/4