What are the solutions to the equation n^2-n=84 ?

The solutions to the equation n^2-n=84 are n=(1+sqrt(337))/2 ,n=(1-sqrt(337))/2

Find the zeros of n^2-n=84

The zeros of n^2-n=84 are n=(1+sqrt(337))/2 ,n=(1-sqrt(337))/2

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n^{2} - n = 84

n = \frac{1 + 337}{2}, n = \frac{1 - 337}{2}

Decimal Notation

n = 9.67877 \dots, n = - 8.67877 \dots

Solution steps

n^{2} - n = 84

Move

84

to the left side

n^{2} - n - 84 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 84 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 84) = 337

n_{1, 2} = \frac{- ( - 1 ) \pm 337}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 337}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 337}{2 \cdot 1}

n = \frac{- ( - 1 ) + 337}{2 \cdot 1} : \frac{1 + 337}{2}

n = \frac{- ( - 1 ) - 337}{2 \cdot 1} : \frac{1 - 337}{2}

The solutions to the quadratic equation are:

n = \frac{1 + 337}{2}, n = \frac{1 - 337}{2}

n^{2} - n = 56

X^{2} + 6 X + 9 = 0

15 x^{2} + 8 x - 12 = 0

0 = 40 - 5 t - 5 t^{2}

(6 x + 4)^{2} = 16

What are the solutions to the equation n^2-n=84 ?
The solutions to the equation n^2-n=84 are n=(1+sqrt(337))/2 ,n=(1-sqrt(337))/2
Find the zeros of n^2-n=84
The zeros of n^2-n=84 are n=(1+sqrt(337))/2 ,n=(1-sqrt(337))/2