What are the solutions to the equation x^2+12x=-28 ?

The solutions to the equation x^2+12x=-28 are x=2(sqrt(2)-3),x=-2(3+sqrt(2))

Find the zeros of x^2+12x=-28

The zeros of x^2+12x=-28 are x=2(sqrt(2)-3),x=-2(3+sqrt(2))

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x^{2} + 12 x = - 28

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

Solution steps

x^{2} + 12 x = - 28

Move

28

to the left side

x^{2} + 12 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 28 = 42

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

Separate the solutions

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

x = \frac{- 12 + 4 2}{2 \cdot 1} : 2 (2 - 3)

x = \frac{- 12 - 4 2}{2 \cdot 1} : - 2 (3 + 2)

The solutions to the quadratic equation are:

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

7 - 3 x > 0

u^{2} + 21 uv + 20 v^{2}

1 6^{- \frac{1}{2}}

(\frac{11}{16})^{2}

4 - 2 x < 5 - 4 x

What are the solutions to the equation x^2+12x=-28 ?
The solutions to the equation x^2+12x=-28 are x=2(sqrt(2)-3),x=-2(3+sqrt(2))
Find the zeros of x^2+12x=-28
The zeros of x^2+12x=-28 are x=2(sqrt(2)-3),x=-2(3+sqrt(2))