What are the solutions to the equation 9n^2-3n-8=-10 ?

The solutions to the equation 9n^2-3n-8=-10 are n= 1/6+i(sqrt(7))/6 ,n= 1/6-i(sqrt(7))/6

Find the zeros of 9n^2-3n-8=-10

The zeros of 9n^2-3n-8=-10 are n= 1/6+i(sqrt(7))/6 ,n= 1/6-i(sqrt(7))/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

9 n^{2} - 3 n - 8 = - 10

n = \frac{1}{6} + i \frac{7}{6}, n = \frac{1}{6} - i \frac{7}{6}

Solution steps

9 n^{2} - 3 n - 8 = - 10

Move

10

to the left side

9 n^{2} - 3 n + 2 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 \cdot 2}{2 \cdot 9}

Simplify

(- 3)^{2} - 4 \cdot 9 \cdot 2 : 37 i

n_{1, 2} = \frac{- ( - 3 ) \pm 3 7 i}{2 \cdot 9}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 3 7 i}{2 \cdot 9}, n_{2} = \frac{- ( - 3 ) - 3 7 i}{2 \cdot 9}

n = \frac{- ( - 3 ) + 3 7 i}{2 \cdot 9} : \frac{1}{6} + i \frac{7}{6}

n = \frac{- ( - 3 ) - 3 7 i}{2 \cdot 9} : \frac{1}{6} - i \frac{7}{6}

The solutions to the quadratic equation are:

n = \frac{1}{6} + i \frac{7}{6}, n = \frac{1}{6} - i \frac{7}{6}

x^{2} - 12 x + 1 = 0

x^{2} + 6 x - 10 = 30

2 x^{2} = 30

5 v^{2} + 18 v + 12 = - 3

12 x^{2} - 276 x + 1080 = 0

What are the solutions to the equation 9n^2-3n-8=-10 ?
The solutions to the equation 9n^2-3n-8=-10 are n= 1/6+i(sqrt(7))/6 ,n= 1/6-i(sqrt(7))/6
Find the zeros of 9n^2-3n-8=-10
The zeros of 9n^2-3n-8=-10 are n= 1/6+i(sqrt(7))/6 ,n= 1/6-i(sqrt(7))/6