What are the solutions to the equation j^2-j=9 ?

The solutions to the equation j^2-j=9 are j=(1+sqrt(37))/2 ,j=(1-sqrt(37))/2

Find the zeros of j^2-j=9

The zeros of j^2-j=9 are j=(1+sqrt(37))/2 ,j=(1-sqrt(37))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

j^{2} - j = 9

j = \frac{1 + 37}{2}, j = \frac{1 - 37}{2}

Decimal Notation

j = 3.54138 \dots, j = - 2.54138 \dots

Solution steps

j^{2} - j = 9

Move

9

to the left side

j^{2} - j - 9 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 9) = 37

j_{1, 2} = \frac{- ( - 1 ) \pm 37}{2 \cdot 1}

Separate the solutions

j_{1} = \frac{- ( - 1 ) + 37}{2 \cdot 1}, j_{2} = \frac{- ( - 1 ) - 37}{2 \cdot 1}

j = \frac{- ( - 1 ) + 37}{2 \cdot 1} : \frac{1 + 37}{2}

j = \frac{- ( - 1 ) - 37}{2 \cdot 1} : \frac{1 - 37}{2}

The solutions to the quadratic equation are:

j = \frac{1 + 37}{2}, j = \frac{1 - 37}{2}

7 + 4 ∣ 1 - 5 x ∣ = 47

- 0.7 + \frac{2}{8}

- 4 (5 + w) > 24

- 3 r > 33

3 (3) + y = 10

What are the solutions to the equation j^2-j=9 ?
The solutions to the equation j^2-j=9 are j=(1+sqrt(37))/2 ,j=(1-sqrt(37))/2
Find the zeros of j^2-j=9
The zeros of j^2-j=9 are j=(1+sqrt(37))/2 ,j=(1-sqrt(37))/2