What are the solutions to the equation x^2+8x=-4 ?

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

Find the zeros of x^2+8x=-4

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

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x^{2} + 8 x = - 4

x = 2 (3 - 2), x = - 2 (2 + 3)

Decimal Notation

x = - 0.53589 \dots, x = - 7.46410 \dots

Solution steps

x^{2} + 8 x = - 4

Move

4

to the left side

x^{2} + 8 x + 4 = 0

Solve with the quadratic formula

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

8^{2} - 4 \cdot 1 \cdot 4 = 43

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

Separate the solutions

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

x = \frac{- 8 + 4 3}{2 \cdot 1} : 2 (3 - 2)

x = \frac{- 8 - 4 3}{2 \cdot 1} : - 2 (2 + 3)

The solutions to the quadratic equation are:

x = 2 (3 - 2), x = - 2 (2 + 3)

x^{2} + 10 x - 36 = 0

2 x^{2} - 6 x = 20

3 x^{2} - 21 = 4 - x^{2}

5 x^{2} = 4

(5 x - 2) (x + 4) = 0

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