What are the solutions to the equation 2x^2+5x+6=0 ?

The solutions to the equation 2x^2+5x+6=0 are x=-5/4+i(sqrt(23))/4 ,x=-5/4-i(sqrt(23))/4

Find the zeros of 2x^2+5x+6=0

The zeros of 2x^2+5x+6=0 are x=-5/4+i(sqrt(23))/4 ,x=-5/4-i(sqrt(23))/4

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2 x^{2} + 5 x + 6 = 0

x = - \frac{5}{4} + i \frac{23}{4}, x = - \frac{5}{4} - i \frac{23}{4}

Solution steps

2 x^{2} + 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

Simplify

5^{2} - 4 \cdot 2 \cdot 6 : 23 i

x_{1, 2} = \frac{- 5 \pm 23 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 23 i}{2 \cdot 2}, x_{2} = \frac{- 5 - 23 i}{2 \cdot 2}

x = \frac{- 5 + 23 i}{2 \cdot 2} : - \frac{5}{4} + i \frac{23}{4}

x = \frac{- 5 - 23 i}{2 \cdot 2} : - \frac{5}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

x = - \frac{5}{4} + i \frac{23}{4}, x = - \frac{5}{4} - i \frac{23}{4}

5 x^{2} + 11 x + 2 \leq 0

\frac{1}{1 + e ^{- 3}}

x^{2} + (1 - x) (1 - 3 x) = x

a (a - 5) < 2 (a - 5)

\frac{2}{6}

What are the solutions to the equation 2x^2+5x+6=0 ?
The solutions to the equation 2x^2+5x+6=0 are x=-5/4+i(sqrt(23))/4 ,x=-5/4-i(sqrt(23))/4
Find the zeros of 2x^2+5x+6=0
The zeros of 2x^2+5x+6=0 are x=-5/4+i(sqrt(23))/4 ,x=-5/4-i(sqrt(23))/4