What are the solutions to the equation x^2+4x=28 ?

The solutions to the equation x^2+4x=28 are x=2(2sqrt(2)-1),x=-2(1+2sqrt(2))

Find the zeros of x^2+4x=28

The zeros of x^2+4x=28 are x=2(2sqrt(2)-1),x=-2(1+2sqrt(2))

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x^{2} + 4 x = 28

x = 2 (22 - 1), x = - 2 (1 + 22)

Decimal Notation

x = 3.65685 \dots, x = - 7.65685 \dots

Solution steps

x^{2} + 4 x = 28

Move

28

to the left side

x^{2} + 4 x - 28 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 28) = 82

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

Separate the solutions

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

x = \frac{- 4 + 8 2}{2 \cdot 1} : 2 (22 - 1)

x = \frac{- 4 - 8 2}{2 \cdot 1} : - 2 (1 + 22)

The solutions to the quadratic equation are:

x = 2 (22 - 1), x = - 2 (1 + 22)

(66 c + 99 d) \cdot \frac{1}{3}

2 x^{3} - x^{2} - x

(x + 8) (x - 9) = 4

\frac{i}{1 + i}

x - \frac{x}{3} + \frac{39}{8} = \frac{x}{4} - \frac{2 x}{6}

What are the solutions to the equation x^2+4x=28 ?
The solutions to the equation x^2+4x=28 are x=2(2sqrt(2)-1),x=-2(1+2sqrt(2))
Find the zeros of x^2+4x=28
The zeros of x^2+4x=28 are x=2(2sqrt(2)-1),x=-2(1+2sqrt(2))