What are the solutions to the equation 4x^2+3x+5=0 ?

The solutions to the equation 4x^2+3x+5=0 are x=-3/8+i(sqrt(71))/8 ,x=-3/8-i(sqrt(71))/8

Find the zeros of 4x^2+3x+5=0

The zeros of 4x^2+3x+5=0 are x=-3/8+i(sqrt(71))/8 ,x=-3/8-i(sqrt(71))/8

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

x = - \frac{3}{8} + i \frac{71}{8}, x = - \frac{3}{8} - i \frac{71}{8}

Solution steps

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

Solve with the quadratic formula

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

Simplify

3^{2} - 4 \cdot 4 \cdot 5 : 71 i

x_{1, 2} = \frac{- 3 \pm 71 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 71 i}{2 \cdot 4}, x_{2} = \frac{- 3 - 71 i}{2 \cdot 4}

x = \frac{- 3 + 71 i}{2 \cdot 4} : - \frac{3}{8} + i \frac{71}{8}

x = \frac{- 3 - 71 i}{2 \cdot 4} : - \frac{3}{8} - i \frac{71}{8}

The solutions to the quadratic equation are:

x = - \frac{3}{8} + i \frac{71}{8}, x = - \frac{3}{8} - i \frac{71}{8}

10 x^{3} + 30 x^{2} + 18 x

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63 (b - 1) + 7 = 7 (9 b - 8)

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

\frac{13 !}{5 ! 8 !}

What are the solutions to the equation 4x^2+3x+5=0 ?
The solutions to the equation 4x^2+3x+5=0 are x=-3/8+i(sqrt(71))/8 ,x=-3/8-i(sqrt(71))/8
Find the zeros of 4x^2+3x+5=0
The zeros of 4x^2+3x+5=0 are x=-3/8+i(sqrt(71))/8 ,x=-3/8-i(sqrt(71))/8