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

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

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

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

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

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

Solution steps

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

Move

36 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

3 6^{2} - 4 \cdot 9 \cdot 38 : 62 i

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

Separate the solutions

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

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

x = \frac{- 36 - 6 2 i}{2 \cdot 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}

- 6 f + 3 < 51

5^{- 3} = \frac{1}{125}

\frac{14 x ^{4} y ^{3}}{21 x ^{8} y ^{5}}

(a^{\frac{2}{3}})^{- 1}

166.72 = - k (- 0.037)

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