What are the solutions to the equation 7j^2-3j=-j+9 ?

The solutions to the equation 7j^2-3j=-j+9 are j= 9/7 ,j=-1

Find the zeros of 7j^2-3j=-j+9

The zeros of 7j^2-3j=-j+9 are j= 9/7 ,j=-1

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7 j^{2} - 3 j = - j + 9

j = \frac{9}{7}, j = - 1

Decimal Notation

j = 1.28571 \dots, j = - 1

Solution steps

7 j^{2} - 3 j = - j + 9

Move

9

to the left side

7 j^{2} - 3 j - 9 = - j

Move

j

to the left side

7 j^{2} - 2 j - 9 = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 7 ( - 9 )}{2 \cdot 7}

(- 2)^{2} - 4 \cdot 7 (- 9) = 16

j_{1, 2} = \frac{- ( - 2 ) \pm 16}{2 \cdot 7}

Separate the solutions

j_{1} = \frac{- ( - 2 ) + 16}{2 \cdot 7}, j_{2} = \frac{- ( - 2 ) - 16}{2 \cdot 7}

j = \frac{- ( - 2 ) + 16}{2 \cdot 7} : \frac{9}{7}

j = \frac{- ( - 2 ) - 16}{2 \cdot 7} : - 1

The solutions to the quadratic equation are:

j = \frac{9}{7}, j = - 1

x^{2} = 7 mod 19

40 - 13 x = - x^{2}

4 x^{2} + x - \frac{1}{4} = 0

7^{2} = 2^{2} + x^{2}

(2 x + 1) (5 x + 2) = (2 x - 2) (x - 2)

What are the solutions to the equation 7j^2-3j=-j+9 ?
The solutions to the equation 7j^2-3j=-j+9 are j= 9/7 ,j=-1
Find the zeros of 7j^2-3j=-j+9
The zeros of 7j^2-3j=-j+9 are j= 9/7 ,j=-1