What are the solutions to the equation 2x^2+3x+6=0 ?

The solutions to the equation 2x^2+3x+6=0 are x=-3/4+i(sqrt(39))/4 ,x=-3/4-i(sqrt(39))/4

Find the zeros of 2x^2+3x+6=0

The zeros of 2x^2+3x+6=0 are x=-3/4+i(sqrt(39))/4 ,x=-3/4-i(sqrt(39))/4

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2 x^{2} + 3 x + 6 = 0

x = - \frac{3}{4} + i \frac{39}{4}, x = - \frac{3}{4} - i \frac{39}{4}

Solution steps

2 x^{2} + 3 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

Simplify

3^{2} - 4 \cdot 2 \cdot 6 : 39 i

x_{1, 2} = \frac{- 3 \pm 39 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 39 i}{2 \cdot 2}, x_{2} = \frac{- 3 - 39 i}{2 \cdot 2}

x = \frac{- 3 + 39 i}{2 \cdot 2} : - \frac{3}{4} + i \frac{39}{4}

x = \frac{- 3 - 39 i}{2 \cdot 2} : - \frac{3}{4} - i \frac{39}{4}

The solutions to the quadratic equation are:

x = - \frac{3}{4} + i \frac{39}{4}, x = - \frac{3}{4} - i \frac{39}{4}

3 x^{2} + x + 7 = 0

6 (- 3 x + 1) = - 17 (x + 1)

\frac{\frac{k ^{2}}{4} - 9}{\frac{k}{6} - 1}

x + 4 = 25

- 150 x^{2} + 1.2 x + (- 380) = 0

What are the solutions to the equation 2x^2+3x+6=0 ?
The solutions to the equation 2x^2+3x+6=0 are x=-3/4+i(sqrt(39))/4 ,x=-3/4-i(sqrt(39))/4
Find the zeros of 2x^2+3x+6=0
The zeros of 2x^2+3x+6=0 are x=-3/4+i(sqrt(39))/4 ,x=-3/4-i(sqrt(39))/4