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

The solutions to the equation 2x+x^2=4 are x=-1+sqrt(5),x=-1-sqrt(5)

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

The zeros of 2x+x^2=4 are x=-1+sqrt(5),x=-1-sqrt(5)

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2 x + x^{2} = 4

x = - 1 + 5, x = - 1 - 5

Solution steps

2 x + x^{2} = 4

Move

4

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 4) = 25

x_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 5}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 5}{2 \cdot 1}

x = \frac{- 2 + 2 5}{2 \cdot 1} : - 1 + 5

x = \frac{- 2 - 2 5}{2 \cdot 1} : - 1 - 5

The solutions to the quadratic equation are:

x = - 1 + 5, x = - 1 - 5

x^{2} - 12 x + 36

x, x^{2} + y^{2} - 6 x - 6 y + 18 = 0

(2 x - 3) (5 x - 1) = x^{2} + 5

w^{2} - 5 w - \frac{11}{4} = 0

4 x^{2} + 16 x + 4 = 0

What are the solutions to the equation 2x+x^2=4 ?
The solutions to the equation 2x+x^2=4 are x=-1+sqrt(5),x=-1-sqrt(5)
Find the zeros of 2x+x^2=4
The zeros of 2x+x^2=4 are x=-1+sqrt(5),x=-1-sqrt(5)