What are the solutions to the equation -9x^2+36x=38 ?

The solutions to the equation -9x^2+36x=38 are x=2-i(sqrt(2))/3 ,x=2+i(sqrt(2))/3

Find the zeros of -9x^2+36x=38

The zeros of -9x^2+36x=38 are x=2-i(sqrt(2))/3 ,x=2+i(sqrt(2))/3

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- 9 x^{2} + 36 x = 38

x = 2 - i \frac{2}{3}, x = 2 + i \frac{2}{3}

Solution steps

- 9 x^{2} + 36 x = 38

Move

38

to the left side

- 9 x^{2} + 36 x - 38 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 ( - 9 ) ( - 38 )}{2 ( - 9 )}

Simplify

3 6^{2} - 4 (- 9) (- 38) : 62 i

x_{1, 2} = \frac{- 36 \pm 6 2 i}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- 36 + 6 2 i}{2 ( - 9 )}, x_{2} = \frac{- 36 - 6 2 i}{2 ( - 9 )}

x = \frac{- 36 + 6 2 i}{2 ( - 9 )} : 2 - i \frac{2}{3}

x = \frac{- 36 - 6 2 i}{2 ( - 9 )} : 2 + i \frac{2}{3}

The solutions to the quadratic equation are:

x = 2 - i \frac{2}{3}, x = 2 + i \frac{2}{3}

x^{2} - \frac{1}{3} x - \frac{2}{9} = 0

2 q^{2} + 4 q - 96 = 0

9 x^{2} - 28 = 21

- x^{2} - 1 = - 65

x^{2} + 1 0^{2} = 1 1^{2}

What are the solutions to the equation -9x^2+36x=38 ?
The solutions to the equation -9x^2+36x=38 are x=2-i(sqrt(2))/3 ,x=2+i(sqrt(2))/3
Find the zeros of -9x^2+36x=38
The zeros of -9x^2+36x=38 are x=2-i(sqrt(2))/3 ,x=2+i(sqrt(2))/3