What are the solutions to the equation -7j^2-2=-8j ?

The solutions to the equation -7j^2-2=-8j are j=(4-sqrt(2))/7 ,j=(4+sqrt(2))/7

Find the zeros of -7j^2-2=-8j

The zeros of -7j^2-2=-8j are j=(4-sqrt(2))/7 ,j=(4+sqrt(2))/7

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- 7 j^{2} - 2 = - 8 j

j = \frac{4 - 2}{7}, j = \frac{4 + 2}{7}

Decimal Notation

j = 0.36939 \dots, j = 0.77345 \dots

Solution steps

- 7 j^{2} - 2 = - 8 j

Move

8 j

to the left side

- 7 j^{2} - 2 + 8 j = 0

Write in the standard form

a x^{2} + b x + c = 0

- 7 j^{2} + 8 j - 2 = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 7 ) ( - 2 )}{2 ( - 7 )}

8^{2} - 4 (- 7) (- 2) = 22

j_{1, 2} = \frac{- 8 \pm 2 2}{2 ( - 7 )}

Separate the solutions

j_{1} = \frac{- 8 + 2 2}{2 ( - 7 )}, j_{2} = \frac{- 8 - 2 2}{2 ( - 7 )}

j = \frac{- 8 + 2 2}{2 ( - 7 )} : \frac{4 - 2}{7}

j = \frac{- 8 - 2 2}{2 ( - 7 )} : \frac{4 + 2}{7}

The solutions to the quadratic equation are:

j = \frac{4 - 2}{7}, j = \frac{4 + 2}{7}

1^{- 5}

3 = x + 3 - 5 x

t^{2} - 10 t + 16

(x^{- 2} x^{- 3})^{4}

\frac{x - 2}{2} + \frac{x - 2 ( x - 4 )}{6} = \frac{x}{3}

What are the solutions to the equation -7j^2-2=-8j ?
The solutions to the equation -7j^2-2=-8j are j=(4-sqrt(2))/7 ,j=(4+sqrt(2))/7
Find the zeros of -7j^2-2=-8j
The zeros of -7j^2-2=-8j are j=(4-sqrt(2))/7 ,j=(4+sqrt(2))/7