What are the solutions to the equation 9x^2+9x+52=0 ?

The solutions to the equation 9x^2+9x+52=0 are x=-1/2+i(sqrt(199))/6 ,x=-1/2-i(sqrt(199))/6

Find the zeros of 9x^2+9x+52=0

The zeros of 9x^2+9x+52=0 are x=-1/2+i(sqrt(199))/6 ,x=-1/2-i(sqrt(199))/6

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9 x^{2} + 9 x + 52 = 0

x = - \frac{1}{2} + i \frac{199}{6}, x = - \frac{1}{2} - i \frac{199}{6}

Solution steps

9 x^{2} + 9 x + 52 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 9 \cdot 52}{2 \cdot 9}

Simplify

9^{2} - 4 \cdot 9 \cdot 52 : 3199 i

x_{1, 2} = \frac{- 9 \pm 3 199 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 9 + 3 199 i}{2 \cdot 9}, x_{2} = \frac{- 9 - 3 199 i}{2 \cdot 9}

x = \frac{- 9 + 3 199 i}{2 \cdot 9} : - \frac{1}{2} + i \frac{199}{6}

x = \frac{- 9 - 3 199 i}{2 \cdot 9} : - \frac{1}{2} - i \frac{199}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{199}{6}, x = - \frac{1}{2} - i \frac{199}{6}

(5 a^{3} b)^{4}

∣ x - 1 ∣ < 9

9 - 2 x^{2} = - 17

\frac{0.628 2 ^{2}}{2 \cdot 0.293}

\frac{16}{81} - 1

What are the solutions to the equation 9x^2+9x+52=0 ?
The solutions to the equation 9x^2+9x+52=0 are x=-1/2+i(sqrt(199))/6 ,x=-1/2-i(sqrt(199))/6
Find the zeros of 9x^2+9x+52=0
The zeros of 9x^2+9x+52=0 are x=-1/2+i(sqrt(199))/6 ,x=-1/2-i(sqrt(199))/6