What are the solutions to the equation -7z^2-98z=307 ?

The solutions to the equation -7z^2-98z=307 are z=-(49+6sqrt(7))/7 ,z=-(49-6sqrt(7))/7

Find the zeros of -7z^2-98z=307

The zeros of -7z^2-98z=307 are z=-(49+6sqrt(7))/7 ,z=-(49-6sqrt(7))/7

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- 7 z^{2} - 98 z = 307

z = - \frac{49 + 6 7}{7}, z = - \frac{49 - 6 7}{7}

Solution steps

- 7 z^{2} - 98 z = 307

Move

307

to the left side

- 7 z^{2} - 98 z - 307 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 98 ) \pm ( - 98 ) ^{2} - 4 ( - 7 ) ( - 307 )}{2 ( - 7 )}

(- 98)^{2} - 4 (- 7) (- 307) = 127

z_{1, 2} = \frac{- ( - 98 ) \pm 12 7}{2 ( - 7 )}

Separate the solutions

z_{1} = \frac{- ( - 98 ) + 12 7}{2 ( - 7 )}, z_{2} = \frac{- ( - 98 ) - 12 7}{2 ( - 7 )}

z = \frac{- ( - 98 ) + 12 7}{2 ( - 7 )} : - \frac{49 + 6 7}{7}

z = \frac{- ( - 98 ) - 12 7}{2 ( - 7 )} : - \frac{49 - 6 7}{7}

The solutions to the quadratic equation are:

z = - \frac{49 + 6 7}{7}, z = - \frac{49 - 6 7}{7}

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What are the solutions to the equation -7z^2-98z=307 ?
The solutions to the equation -7z^2-98z=307 are z=-(49+6sqrt(7))/7 ,z=-(49-6sqrt(7))/7
Find the zeros of -7z^2-98z=307
The zeros of -7z^2-98z=307 are z=-(49+6sqrt(7))/7 ,z=-(49-6sqrt(7))/7