What are the solutions to the equation 2x^2+12x+20=0 ?

The solutions to the equation 2x^2+12x+20=0 are x=-3+i,x=-3-i

Find the zeros of 2x^2+12x+20=0

The zeros of 2x^2+12x+20=0 are x=-3+i,x=-3-i

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2 x^{2} + 12 x + 20 = 0

x = - 3 + i, x = - 3 - i

Solution steps

2 x^{2} + 12 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

Simplify

1 2^{2} - 4 \cdot 2 \cdot 20 : 4 i

x_{1, 2} = \frac{- 12 \pm 4 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 4 i}{2 \cdot 2}, x_{2} = \frac{- 12 - 4 i}{2 \cdot 2}

x = \frac{- 12 + 4 i}{2 \cdot 2} : - 3 + i

x = \frac{- 12 - 4 i}{2 \cdot 2} : - 3 - i

The solutions to the quadratic equation are:

x = - 3 + i, x = - 3 - i

- \frac{u}{3} = - 31

x^{2} = \frac{24}{25}

2 \geq \frac{1}{a - 1} > 1

x^{2} + 2 x - 1 < 0

3 \cdot 7

What are the solutions to the equation 2x^2+12x+20=0 ?
The solutions to the equation 2x^2+12x+20=0 are x=-3+i,x=-3-i
Find the zeros of 2x^2+12x+20=0
The zeros of 2x^2+12x+20=0 are x=-3+i,x=-3-i