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

The solutions to the equation x^2+4x=2 are x=-2+sqrt(6),x=-2-sqrt(6)

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

The zeros of x^2+4x=2 are x=-2+sqrt(6),x=-2-sqrt(6)

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x^{2} + 4 x = 2

x = - 2 + 6, x = - 2 - 6

Decimal Notation

x = 0.44948 \dots, x = - 4.44948 \dots

Solution steps

x^{2} + 4 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 2) = 26

x_{1, 2} = \frac{- 4 \pm 2 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 6}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 6}{2 \cdot 1}

x = \frac{- 4 + 2 6}{2 \cdot 1} : - 2 + 6

x = \frac{- 4 - 2 6}{2 \cdot 1} : - 2 - 6

The solutions to the quadratic equation are:

x = - 2 + 6, x = - 2 - 6

- \frac{1}{10} < \frac{1}{2} - \frac{2}{5} x \leq \frac{4}{2}

4 v^{2} + 12 v + 9

(\frac{( 3 x y ^{- 5} ) ^{3}}{( x ^{- 2} y ^{2} ) ^{- 4}})^{- 2}

- 11 \cdot - 11

x - \frac{2 - x}{3} = - \frac{x + 1}{3} + \frac{3}{2}

What are the solutions to the equation x^2+4x=2 ?
The solutions to the equation x^2+4x=2 are x=-2+sqrt(6),x=-2-sqrt(6)
Find the zeros of x^2+4x=2
The zeros of x^2+4x=2 are x=-2+sqrt(6),x=-2-sqrt(6)