What are the solutions to the equation 7x-8x^2=9 ?

The solutions to the equation 7x-8x^2=9 are x= 7/16-i(sqrt(239))/(16),x= 7/16+i(sqrt(239))/(16)

Find the zeros of 7x-8x^2=9

The zeros of 7x-8x^2=9 are x= 7/16-i(sqrt(239))/(16),x= 7/16+i(sqrt(239))/(16)

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7 x - 8 x^{2} = 9

x = \frac{7}{16} - i \frac{239}{16}, x = \frac{7}{16} + i \frac{239}{16}

Solution steps

7 x - 8 x^{2} = 9

Move

9

to the left side

7 x - 8 x^{2} - 9 = 0

Write in the standard form

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

- 8 x^{2} + 7 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 8 ) ( - 9 )}{2 ( - 8 )}

Simplify

7^{2} - 4 (- 8) (- 9) : 239 i

x_{1, 2} = \frac{- 7 \pm 239 i}{2 ( - 8 )}

Separate the solutions

x_{1} = \frac{- 7 + 239 i}{2 ( - 8 )}, x_{2} = \frac{- 7 - 239 i}{2 ( - 8 )}

x = \frac{- 7 + 239 i}{2 ( - 8 )} : \frac{7}{16} - i \frac{239}{16}

x = \frac{- 7 - 239 i}{2 ( - 8 )} : \frac{7}{16} + i \frac{239}{16}

The solutions to the quadratic equation are:

x = \frac{7}{16} - i \frac{239}{16}, x = \frac{7}{16} + i \frac{239}{16}

\frac{x - 6}{9} = - 5

\frac{5}{8} - \frac{1}{8}

2 - \frac{1}{4} x > 0

(x + 1)^{2} = 5

19 t + 20 = 9 t + 80

What are the solutions to the equation 7x-8x^2=9 ?
The solutions to the equation 7x-8x^2=9 are x= 7/16-i(sqrt(239))/(16),x= 7/16+i(sqrt(239))/(16)
Find the zeros of 7x-8x^2=9
The zeros of 7x-8x^2=9 are x= 7/16-i(sqrt(239))/(16),x= 7/16+i(sqrt(239))/(16)