What are the solutions to the equation 3n^2-2n-18=2 ?

The solutions to the equation 3n^2-2n-18=2 are n=(1+sqrt(61))/3 ,n=(1-sqrt(61))/3

Find the zeros of 3n^2-2n-18=2

The zeros of 3n^2-2n-18=2 are n=(1+sqrt(61))/3 ,n=(1-sqrt(61))/3

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3 n^{2} - 2 n - 18 = 2

n = \frac{1 + 61}{3}, n = \frac{1 - 61}{3}

Decimal Notation

n = 2.93674 \dots, n = - 2.27008 \dots

Solution steps

3 n^{2} - 2 n - 18 = 2

Move

2

to the left side

3 n^{2} - 2 n - 20 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 20) = 261

n_{1, 2} = \frac{- ( - 2 ) \pm 2 61}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 2 61}{2 \cdot 3}, n_{2} = \frac{- ( - 2 ) - 2 61}{2 \cdot 3}

n = \frac{- ( - 2 ) + 2 61}{2 \cdot 3} : \frac{1 + 61}{3}

n = \frac{- ( - 2 ) - 2 61}{2 \cdot 3} : \frac{1 - 61}{3}

The solutions to the quadratic equation are:

n = \frac{1 + 61}{3}, n = \frac{1 - 61}{3}

33 x - 33 = 33 x + 25

\frac{b}{2} + \frac{b}{3} - 26 \geq 24 - 48

∣ x + 6 ∣ = 11

2 > \frac{c}{3}

5 ∣ f ∣ = 0

What are the solutions to the equation 3n^2-2n-18=2 ?
The solutions to the equation 3n^2-2n-18=2 are n=(1+sqrt(61))/3 ,n=(1-sqrt(61))/3
Find the zeros of 3n^2-2n-18=2
The zeros of 3n^2-2n-18=2 are n=(1+sqrt(61))/3 ,n=(1-sqrt(61))/3