What are the solutions to the equation 5x^2+20x+32=0 ?

The solutions to the equation 5x^2+20x+32=0 are x=-2+i(2sqrt(15))/5 ,x=-2-i(2sqrt(15))/5

Find the zeros of 5x^2+20x+32=0

The zeros of 5x^2+20x+32=0 are x=-2+i(2sqrt(15))/5 ,x=-2-i(2sqrt(15))/5

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5 x^{2} + 20 x + 32 = 0

x = - 2 + i \frac{2 15}{5}, x = - 2 - i \frac{2 15}{5}

Solution steps

5 x^{2} + 20 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 5 \cdot 32}{2 \cdot 5}

Simplify

2 0^{2} - 4 \cdot 5 \cdot 32 : 415 i

x_{1, 2} = \frac{- 20 \pm 4 15 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 20 + 4 15 i}{2 \cdot 5}, x_{2} = \frac{- 20 - 4 15 i}{2 \cdot 5}

x = \frac{- 20 + 4 15 i}{2 \cdot 5} : - 2 + i \frac{2 15}{5}

x = \frac{- 20 - 4 15 i}{2 \cdot 5} : - 2 - i \frac{2 15}{5}

The solutions to the quadratic equation are:

x = - 2 + i \frac{2 15}{5}, x = - 2 - i \frac{2 15}{5}

2 x^{2} - 7 x + 3

m^{2} - 12 m + 36

x^{3} - 5 x^{2} - 14 x \geq 0

3 x^{2} + π x - 4 < 0

\frac{2 x}{5} > 1

What are the solutions to the equation 5x^2+20x+32=0 ?
The solutions to the equation 5x^2+20x+32=0 are x=-2+i(2sqrt(15))/5 ,x=-2-i(2sqrt(15))/5
Find the zeros of 5x^2+20x+32=0
The zeros of 5x^2+20x+32=0 are x=-2+i(2sqrt(15))/5 ,x=-2-i(2sqrt(15))/5