What are the solutions to the equation 4x^2+3x+1=-2 ?

The solutions to the equation 4x^2+3x+1=-2 are x=-3/8+i(sqrt(39))/8 ,x=-3/8-i(sqrt(39))/8

Find the zeros of 4x^2+3x+1=-2

The zeros of 4x^2+3x+1=-2 are x=-3/8+i(sqrt(39))/8 ,x=-3/8-i(sqrt(39))/8

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4 x^{2} + 3 x + 1 = - 2

x = - \frac{3}{8} + i \frac{39}{8}, x = - \frac{3}{8} - i \frac{39}{8}

Solution steps

4 x^{2} + 3 x + 1 = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- 3 \pm 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 39 i}{2 \cdot 4}, x_{2} = \frac{- 3 - 39 i}{2 \cdot 4}

x = \frac{- 3 + 39 i}{2 \cdot 4} : - \frac{3}{8} + i \frac{39}{8}

x = \frac{- 3 - 39 i}{2 \cdot 4} : - \frac{3}{8} - i \frac{39}{8}

The solutions to the quadratic equation are:

x = - \frac{3}{8} + i \frac{39}{8}, x = - \frac{3}{8} - i \frac{39}{8}

- \frac{5}{2 ( \frac{1}{16} ) ^{\frac{3}{2}}}

10 x + 24 < 16 x + 12

2 - x \leq 1

n^{2} - 14 n + 24

- 2 m - 12 \leq - 24 and 7 m + 8 \geq 22

What are the solutions to the equation 4x^2+3x+1=-2 ?
The solutions to the equation 4x^2+3x+1=-2 are x=-3/8+i(sqrt(39))/8 ,x=-3/8-i(sqrt(39))/8
Find the zeros of 4x^2+3x+1=-2
The zeros of 4x^2+3x+1=-2 are x=-3/8+i(sqrt(39))/8 ,x=-3/8-i(sqrt(39))/8