What are the solutions to the equation 10n^2-n-8=0 ?

The solutions to the equation 10n^2-n-8=0 are n=(1+sqrt(321))/(20),n=(1-sqrt(321))/(20)

Find the zeros of 10n^2-n-8=0

The zeros of 10n^2-n-8=0 are n=(1+sqrt(321))/(20),n=(1-sqrt(321))/(20)

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10 n^{2} - n - 8 = 0

n = \frac{1 + 321}{20}, n = \frac{1 - 321}{20}

Decimal Notation

n = 0.94582 \dots, n = - 0.84582 \dots

Solution steps

10 n^{2} - n - 8 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

(- 1)^{2} - 4 \cdot 10 (- 8) = 321

n_{1, 2} = \frac{- ( - 1 ) \pm 321}{2 \cdot 10}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 321}{2 \cdot 10}, n_{2} = \frac{- ( - 1 ) - 321}{2 \cdot 10}

n = \frac{- ( - 1 ) + 321}{2 \cdot 10} : \frac{1 + 321}{20}

n = \frac{- ( - 1 ) - 321}{2 \cdot 10} : \frac{1 - 321}{20}

The solutions to the quadratic equation are:

n = \frac{1 + 321}{20}, n = \frac{1 - 321}{20}

s, f = {\frac{1 - 2 s ^{2}}{- x + 2}}

x^{2} - 40 x - 800 = 0

6 x^{2} + 120 x + 2070 = 0

x, x^{2} + 2 x y + y^{2} + 2 x - 4 y + 5 = 0

6 x^{2} = 12 + x

What are the solutions to the equation 10n^2-n-8=0 ?
The solutions to the equation 10n^2-n-8=0 are n=(1+sqrt(321))/(20),n=(1-sqrt(321))/(20)
Find the zeros of 10n^2-n-8=0
The zeros of 10n^2-n-8=0 are n=(1+sqrt(321))/(20),n=(1-sqrt(321))/(20)