What are the solutions to the equation x^2+11x+42=0 ?

The solutions to the equation x^2+11x+42=0 are x=-11/2+i(sqrt(47))/2 ,x=-11/2-i(sqrt(47))/2

Find the zeros of x^2+11x+42=0

The zeros of x^2+11x+42=0 are x=-11/2+i(sqrt(47))/2 ,x=-11/2-i(sqrt(47))/2

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x^{2} + 11 x + 42 = 0

x = - \frac{11}{2} + i \frac{47}{2}, x = - \frac{11}{2} - i \frac{47}{2}

Solution steps

x^{2} + 11 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

Simplify

1 1^{2} - 4 \cdot 1 \cdot 42 : 47 i

x_{1, 2} = \frac{- 11 \pm 47 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 11 + 47 i}{2 \cdot 1}, x_{2} = \frac{- 11 - 47 i}{2 \cdot 1}

x = \frac{- 11 + 47 i}{2 \cdot 1} : - \frac{11}{2} + i \frac{47}{2}

x = \frac{- 11 - 47 i}{2 \cdot 1} : - \frac{11}{2} - i \frac{47}{2}

The solutions to the quadratic equation are:

x = - \frac{11}{2} + i \frac{47}{2}, x = - \frac{11}{2} - i \frac{47}{2}

9 x > 3

- \frac{2}{3} ab c + \frac{1}{4} ab c - \frac{1}{12} ab c - \frac{7}{8} ab

3 a > - 6 and 2 a + 4 < 6

\frac{1}{2} (\frac{2}{3})^{2} - 4 (\frac{2}{3}) + 6

x^{3} + 2 x^{2} - 9 x - 18

What are the solutions to the equation x^2+11x+42=0 ?
The solutions to the equation x^2+11x+42=0 are x=-11/2+i(sqrt(47))/2 ,x=-11/2-i(sqrt(47))/2
Find the zeros of x^2+11x+42=0
The zeros of x^2+11x+42=0 are x=-11/2+i(sqrt(47))/2 ,x=-11/2-i(sqrt(47))/2