What are the solutions to the equation 2x^2+5x+11=0 ?

The solutions to the equation 2x^2+5x+11=0 are x=-5/4+i(3sqrt(7))/4 ,x=-5/4-i(3sqrt(7))/4

Find the zeros of 2x^2+5x+11=0

The zeros of 2x^2+5x+11=0 are x=-5/4+i(3sqrt(7))/4 ,x=-5/4-i(3sqrt(7))/4

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2 x^{2} + 5 x + 11 = 0

x = - \frac{5}{4} + i \frac{3 7}{4}, x = - \frac{5}{4} - i \frac{3 7}{4}

Solution steps

2 x^{2} + 5 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 11}{2 \cdot 2}

Simplify

5^{2} - 4 \cdot 2 \cdot 11 : 37 i

x_{1, 2} = \frac{- 5 \pm 3 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 3 7 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 3 7 i}{2 \cdot 2}

x = \frac{- 5 + 3 7 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{3 7}{4}

x = \frac{- 5 - 3 7 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{3 7}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{3 7}{4}, x = - \frac{5}{4} - i \frac{3 7}{4}

- 2 x^{2} + 11 x + 21

(ln (2))^{2}

\frac{1}{\frac{1}{27} + \frac{1}{35} + \frac{1}{52} + \frac{1}{53} + \frac{1}{60}}

\frac{x + 1}{x - 5} < 3

5 x + 2 \leq 17

What are the solutions to the equation 2x^2+5x+11=0 ?
The solutions to the equation 2x^2+5x+11=0 are x=-5/4+i(3sqrt(7))/4 ,x=-5/4-i(3sqrt(7))/4
Find the zeros of 2x^2+5x+11=0
The zeros of 2x^2+5x+11=0 are x=-5/4+i(3sqrt(7))/4 ,x=-5/4-i(3sqrt(7))/4