What are the solutions to the equation x^2+8x=-28 ?

The solutions to the equation x^2+8x=-28 are x=-4+2sqrt(3)i,x=-4-2sqrt(3)i

Find the zeros of x^2+8x=-28

The zeros of x^2+8x=-28 are x=-4+2sqrt(3)i,x=-4-2sqrt(3)i

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x^{2} + 8 x = - 28

x = - 4 + 23 i, x = - 4 - 23 i

Solution steps

x^{2} + 8 x = - 28

Move

28

to the left side

x^{2} + 8 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 28 : 43 i

x_{1, 2} = \frac{- 8 \pm 4 3 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 3 i}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 3 i}{2 \cdot 1}

x = \frac{- 8 + 4 3 i}{2 \cdot 1} : - 4 + 23 i

x = \frac{- 8 - 4 3 i}{2 \cdot 1} : - 4 - 23 i

The solutions to the quadratic equation are:

x = - 4 + 23 i, x = - 4 - 23 i

x, x^{2} + x y + y^{2} = 1

s^{2} = 144

(3 x + 1)^{2} = 9

x^{2} + 245 = 0

5 x^{2} + 8 x + 2

What are the solutions to the equation x^2+8x=-28 ?
The solutions to the equation x^2+8x=-28 are x=-4+2sqrt(3)i,x=-4-2sqrt(3)i
Find the zeros of x^2+8x=-28
The zeros of x^2+8x=-28 are x=-4+2sqrt(3)i,x=-4-2sqrt(3)i