What are the solutions to the equation 4x^2+20x+31=0 ?

The solutions to the equation 4x^2+20x+31=0 are x=-5/2+isqrt(3/2),x=-5/2-isqrt(3/2)

Find the zeros of 4x^2+20x+31=0

The zeros of 4x^2+20x+31=0 are x=-5/2+isqrt(3/2),x=-5/2-isqrt(3/2)

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4 x^{2} + 20 x + 31 = 0

x = - \frac{5}{2} + i \frac{3}{2}, x = - \frac{5}{2} - i \frac{3}{2}

Solution steps

4 x^{2} + 20 x + 31 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 31}{2 \cdot 4}

Simplify

2 0^{2} - 4 \cdot 4 \cdot 31 : 46 i

x_{1, 2} = \frac{- 20 \pm 4 6 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 4 6 i}{2 \cdot 4}, x_{2} = \frac{- 20 - 4 6 i}{2 \cdot 4}

x = \frac{- 20 + 4 6 i}{2 \cdot 4} : - \frac{5}{2} + i \frac{3}{2}

x = \frac{- 20 - 4 6 i}{2 \cdot 4} : - \frac{5}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i \frac{3}{2}, x = - \frac{5}{2} - i \frac{3}{2}

\frac{X}{7} - 1 = \frac{X}{6}

24 > - 4 x > 12

∣ 3 y + 8 ∣ + 2 < 0

x^{2} - 0.632 x + 0.1 = 0

n^{2} - 1

What are the solutions to the equation 4x^2+20x+31=0 ?
The solutions to the equation 4x^2+20x+31=0 are x=-5/2+isqrt(3/2),x=-5/2-isqrt(3/2)
Find the zeros of 4x^2+20x+31=0
The zeros of 4x^2+20x+31=0 are x=-5/2+isqrt(3/2),x=-5/2-isqrt(3/2)