What are the solutions to the equation 2x^2+40x=-254 ?

The solutions to the equation 2x^2+40x=-254 are x=-10+3sqrt(3)i,x=-10-3sqrt(3)i

Find the zeros of 2x^2+40x=-254

The zeros of 2x^2+40x=-254 are x=-10+3sqrt(3)i,x=-10-3sqrt(3)i

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2 x^{2} + 40 x = - 254

x = - 10 + 33 i, x = - 10 - 33 i

Solution steps

2 x^{2} + 40 x = - 254

Move

254

to the left side

2 x^{2} + 40 x + 254 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 2 \cdot 254}{2 \cdot 2}

Simplify

4 0^{2} - 4 \cdot 2 \cdot 254 : 123 i

x_{1, 2} = \frac{- 40 \pm 12 3 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 40 + 12 3 i}{2 \cdot 2}, x_{2} = \frac{- 40 - 12 3 i}{2 \cdot 2}

x = \frac{- 40 + 12 3 i}{2 \cdot 2} : - 10 + 33 i

x = \frac{- 40 - 12 3 i}{2 \cdot 2} : - 10 - 33 i

The solutions to the quadratic equation are:

x = - 10 + 33 i, x = - 10 - 33 i

v + 2 = v - 2

2 π 4

- 2 x = 8

6 b^{4} + 2 b

4 x^{2} + 12

What are the solutions to the equation 2x^2+40x=-254 ?
The solutions to the equation 2x^2+40x=-254 are x=-10+3sqrt(3)i,x=-10-3sqrt(3)i
Find the zeros of 2x^2+40x=-254
The zeros of 2x^2+40x=-254 are x=-10+3sqrt(3)i,x=-10-3sqrt(3)i