What are the solutions to the equation 9x^2+5x+1=0 ?

The solutions to the equation 9x^2+5x+1=0 are x=-5/18+i(sqrt(11))/(18),x=-5/18-i(sqrt(11))/(18)

Find the zeros of 9x^2+5x+1=0

The zeros of 9x^2+5x+1=0 are x=-5/18+i(sqrt(11))/(18),x=-5/18-i(sqrt(11))/(18)

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9 x^{2} + 5 x + 1 = 0

x = - \frac{5}{18} + i \frac{11}{18}, x = - \frac{5}{18} - i \frac{11}{18}

Solution steps

9 x^{2} + 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

Simplify

5^{2} - 4 \cdot 9 \cdot 1 : 11 i

x_{1, 2} = \frac{- 5 \pm 11 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 5 + 11 i}{2 \cdot 9}, x_{2} = \frac{- 5 - 11 i}{2 \cdot 9}

x = \frac{- 5 + 11 i}{2 \cdot 9} : - \frac{5}{18} + i \frac{11}{18}

x = \frac{- 5 - 11 i}{2 \cdot 9} : - \frac{5}{18} - i \frac{11}{18}

The solutions to the quadratic equation are:

x = - \frac{5}{18} + i \frac{11}{18}, x = - \frac{5}{18} - i \frac{11}{18}

(x - 2)^{2} + (x - 4)^{2} = x^{2}

t^{2} + 10 t + 9 = 0

11 x^{2} - 153 x - 180 = 0

a^{2} + a + 1 = 0

\frac{c ^{2}}{8} - 3 = 2

What are the solutions to the equation 9x^2+5x+1=0 ?
The solutions to the equation 9x^2+5x+1=0 are x=-5/18+i(sqrt(11))/(18),x=-5/18-i(sqrt(11))/(18)
Find the zeros of 9x^2+5x+1=0
The zeros of 9x^2+5x+1=0 are x=-5/18+i(sqrt(11))/(18),x=-5/18-i(sqrt(11))/(18)