What are the solutions to the equation 2x^2+20x=-58 ?

The solutions to the equation 2x^2+20x=-58 are x=-5+2i,x=-5-2i

Find the zeros of 2x^2+20x=-58

The zeros of 2x^2+20x=-58 are x=-5+2i,x=-5-2i

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2 x^{2} + 20 x = - 58

x = - 5 + 2 i, x = - 5 - 2 i

Solution steps

2 x^{2} + 20 x = - 58

Move

58

to the left side

2 x^{2} + 20 x + 58 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 \cdot 58}{2 \cdot 2}

Simplify

2 0^{2} - 4 \cdot 2 \cdot 58 : 8 i

x_{1, 2} = \frac{- 20 \pm 8 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 20 + 8 i}{2 \cdot 2}, x_{2} = \frac{- 20 - 8 i}{2 \cdot 2}

x = \frac{- 20 + 8 i}{2 \cdot 2} : - 5 + 2 i

x = \frac{- 20 - 8 i}{2 \cdot 2} : - 5 - 2 i

The solutions to the quadratic equation are:

x = - 5 + 2 i, x = - 5 - 2 i

(x + 3)^{2} - 1 = 5

(2 x - 3)^{2} - 5 = 5

(2 x - 3)^{2} - 5 = 4

4 x^{2} + 13 x = - 3

2 n^{2} - 8 n = 3

What are the solutions to the equation 2x^2+20x=-58 ?
The solutions to the equation 2x^2+20x=-58 are x=-5+2i,x=-5-2i
Find the zeros of 2x^2+20x=-58
The zeros of 2x^2+20x=-58 are x=-5+2i,x=-5-2i