What are the solutions to the equation x^2+20/3 x+64=0 ?

The solutions to the equation x^2+20/3 x+64=0 are x=-10/3+i(2sqrt(119))/3 ,x=-10/3-i(2sqrt(119))/3

Find the zeros of x^2+20/3 x+64=0

The zeros of x^2+20/3 x+64=0 are x=-10/3+i(2sqrt(119))/3 ,x=-10/3-i(2sqrt(119))/3

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x^{2} + \frac{20}{3} x + 64 = 0

x = - \frac{10}{3} + i \frac{2 119}{3}, x = - \frac{10}{3} - i \frac{2 119}{3}

Solution steps

x^{2} + \frac{20}{3} x + 64 = 0

Multiply both sides by

3

3 x^{2} + 20 x + 192 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 3 \cdot 192}{2 \cdot 3}

Simplify

2 0^{2} - 4 \cdot 3 \cdot 192 : 4119 i

x_{1, 2} = \frac{- 20 \pm 4 119 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 20 + 4 119 i}{2 \cdot 3}, x_{2} = \frac{- 20 - 4 119 i}{2 \cdot 3}

x = \frac{- 20 + 4 119 i}{2 \cdot 3} : - \frac{10}{3} + i \frac{2 119}{3}

x = \frac{- 20 - 4 119 i}{2 \cdot 3} : - \frac{10}{3} - i \frac{2 119}{3}

The solutions to the quadratic equation are:

x = - \frac{10}{3} + i \frac{2 119}{3}, x = - \frac{10}{3} - i \frac{2 119}{3}

x (x + 2) = 224

y, y^{2} = (x + 1) (x - 1) (x - 2)

\frac{3 x ^{2} - 248 x}{4} + 961 = 0

t^{2} - 6 t + 1 = 0

- 0.12 x^{2} + 10 x + 800 = 0

What are the solutions to the equation x^2+20/3 x+64=0 ?
The solutions to the equation x^2+20/3 x+64=0 are x=-10/3+i(2sqrt(119))/3 ,x=-10/3-i(2sqrt(119))/3
Find the zeros of x^2+20/3 x+64=0
The zeros of x^2+20/3 x+64=0 are x=-10/3+i(2sqrt(119))/3 ,x=-10/3-i(2sqrt(119))/3