What are the solutions to the equation 4x^2+5x+2=0 ?

The solutions to the equation 4x^2+5x+2=0 are x=-5/8+i(sqrt(7))/8 ,x=-5/8-i(sqrt(7))/8

Find the zeros of 4x^2+5x+2=0

The zeros of 4x^2+5x+2=0 are x=-5/8+i(sqrt(7))/8 ,x=-5/8-i(sqrt(7))/8

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

x = - \frac{5}{8} + i \frac{7}{8}, x = - \frac{5}{8} - i \frac{7}{8}

Solution steps

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

Solve with the quadratic formula

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

Simplify

5^{2} - 4 \cdot 4 \cdot 2 : 7 i

x_{1, 2} = \frac{- 5 \pm 7 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 5 + 7 i}{2 \cdot 4}, x_{2} = \frac{- 5 - 7 i}{2 \cdot 4}

x = \frac{- 5 + 7 i}{2 \cdot 4} : - \frac{5}{8} + i \frac{7}{8}

x = \frac{- 5 - 7 i}{2 \cdot 4} : - \frac{5}{8} - i \frac{7}{8}

The solutions to the quadratic equation are:

x = - \frac{5}{8} + i \frac{7}{8}, x = - \frac{5}{8} - i \frac{7}{8}

8 x + 74 + 2 x + 56 = 180

x^{2} - 2 x - 2 p - p^{2}

(2 x^{3} y^{2})^{- 2} (\frac{1}{3} x^{3} y)^{2}

1 + 8 a^{3}

- x^{2} + 1 > x + 1

What are the solutions to the equation 4x^2+5x+2=0 ?
The solutions to the equation 4x^2+5x+2=0 are x=-5/8+i(sqrt(7))/8 ,x=-5/8-i(sqrt(7))/8
Find the zeros of 4x^2+5x+2=0
The zeros of 4x^2+5x+2=0 are x=-5/8+i(sqrt(7))/8 ,x=-5/8-i(sqrt(7))/8