What are the solutions to the equation 2x(5+x)=3+x^2 ?

The solutions to the equation 2x(5+x)=3+x^2 are x=-5+2sqrt(7),x=-5-2sqrt(7)

Find the zeros of 2x(5+x)=3+x^2

The zeros of 2x(5+x)=3+x^2 are x=-5+2sqrt(7),x=-5-2sqrt(7)

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

x = - 5 + 27, x = - 5 - 27

Decimal Notation

x = 0.29150 \dots, x = - 10.29150 \dots

Solution steps

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

Expand

2 x (5 + x) : 10 x + 2 x^{2}

10 x + 2 x^{2} = 3 + x^{2}

Move

x^{2}

to the left side

10 x + x^{2} = 3

Move

3

to the left side

10 x + x^{2} - 3 = 0

Write in the standard form

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

x^{2} + 10 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 3) = 47

x_{1, 2} = \frac{- 10 \pm 4 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 4 7}{2 \cdot 1}, x_{2} = \frac{- 10 - 4 7}{2 \cdot 1}

x = \frac{- 10 + 4 7}{2 \cdot 1} : - 5 + 27

x = \frac{- 10 - 4 7}{2 \cdot 1} : - 5 - 27

The solutions to the quadratic equation are:

x = - 5 + 27, x = - 5 - 27

- 64 x (- 16 x^{2} + 1) \geq 0

s^{3} - 64 t^{3}

∣ 7 x + 1 ∣ - 1 = 11

- 5 z^{2} - 3 z - 11 = - 6 z^{2}

x^{2} = 128

What are the solutions to the equation 2x(5+x)=3+x^2 ?
The solutions to the equation 2x(5+x)=3+x^2 are x=-5+2sqrt(7),x=-5-2sqrt(7)
Find the zeros of 2x(5+x)=3+x^2
The zeros of 2x(5+x)=3+x^2 are x=-5+2sqrt(7),x=-5-2sqrt(7)