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

The solutions to the equation 2x^2+3x+2=-3 are x=-3/4+i(sqrt(31))/4 ,x=-3/4-i(sqrt(31))/4

Find the zeros of 2x^2+3x+2=-3

The zeros of 2x^2+3x+2=-3 are x=-3/4+i(sqrt(31))/4 ,x=-3/4-i(sqrt(31))/4

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2 x^{2} + 3 x + 2 = - 3

x = - \frac{3}{4} + i \frac{31}{4}, x = - \frac{3}{4} - i \frac{31}{4}

Solution steps

2 x^{2} + 3 x + 2 = - 3

Move

3

to the left side

2 x^{2} + 3 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

3^{2} - 4 \cdot 2 \cdot 5 : 31 i

x_{1, 2} = \frac{- 3 \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 31 i}{2 \cdot 2}, x_{2} = \frac{- 3 - 31 i}{2 \cdot 2}

x = \frac{- 3 + 31 i}{2 \cdot 2} : - \frac{3}{4} + i \frac{31}{4}

x = \frac{- 3 - 31 i}{2 \cdot 2} : - \frac{3}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = - \frac{3}{4} + i \frac{31}{4}, x = - \frac{3}{4} - i \frac{31}{4}

3^{\frac{2}{5}} \cdot 3^{(- \frac{7}{5})}

\frac{9 !}{7 ! 2 !}

1 > \frac{b}{- 2} - 1

(\frac{2}{x ^{3}})^{3}

- 2 π \leq x \leq 2 π

What are the solutions to the equation 2x^2+3x+2=-3 ?
The solutions to the equation 2x^2+3x+2=-3 are x=-3/4+i(sqrt(31))/4 ,x=-3/4-i(sqrt(31))/4
Find the zeros of 2x^2+3x+2=-3
The zeros of 2x^2+3x+2=-3 are x=-3/4+i(sqrt(31))/4 ,x=-3/4-i(sqrt(31))/4