What are the solutions to the equation 6x^2+5x+4=0 ?

The solutions to the equation 6x^2+5x+4=0 are x=-5/12+i(sqrt(71))/(12),x=-5/12-i(sqrt(71))/(12)

Find the zeros of 6x^2+5x+4=0

The zeros of 6x^2+5x+4=0 are x=-5/12+i(sqrt(71))/(12),x=-5/12-i(sqrt(71))/(12)

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

x = - \frac{5}{12} + i \frac{71}{12}, x = - \frac{5}{12} - i \frac{71}{12}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 \cdot 4}{2 \cdot 6}

Simplify

5^{2} - 4 \cdot 6 \cdot 4 : 71 i

x_{1, 2} = \frac{- 5 \pm 71 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 5 + 71 i}{2 \cdot 6}, x_{2} = \frac{- 5 - 71 i}{2 \cdot 6}

x = \frac{- 5 + 71 i}{2 \cdot 6} : - \frac{5}{12} + i \frac{71}{12}

x = \frac{- 5 - 71 i}{2 \cdot 6} : - \frac{5}{12} - i \frac{71}{12}

The solutions to the quadratic equation are:

x = - \frac{5}{12} + i \frac{71}{12}, x = - \frac{5}{12} - i \frac{71}{12}

x^{2} - 18 x + 57 = 0

4 x + 8 + 5 \leq 3

3 x - 5 - 4 - x > 1

x^{2} = 21 + 4 x

x^{2} + 10 x y - 24 y^{2}

What are the solutions to the equation 6x^2+5x+4=0 ?
The solutions to the equation 6x^2+5x+4=0 are x=-5/12+i(sqrt(71))/(12),x=-5/12-i(sqrt(71))/(12)
Find the zeros of 6x^2+5x+4=0
The zeros of 6x^2+5x+4=0 are x=-5/12+i(sqrt(71))/(12),x=-5/12-i(sqrt(71))/(12)