What are the solutions to the equation 2x(x+2)+2x(x+1)=x^2+4 ?

The solutions to the equation 2x(x+2)+2x(x+1)=x^2+4 are x=(-3+sqrt(21))/3 ,x=-(3+sqrt(21))/3

Find the zeros of 2x(x+2)+2x(x+1)=x^2+4

The zeros of 2x(x+2)+2x(x+1)=x^2+4 are x=(-3+sqrt(21))/3 ,x=-(3+sqrt(21))/3

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2 x (x + 2) + 2 x (x + 1) = x^{2} + 4

x = \frac{- 3 + 21}{3}, x = - \frac{3 + 21}{3}

Decimal Notation

x = 0.52752 \dots, x = - 2.52752 \dots

Solution steps

2 x (x + 2) + 2 x (x + 1) = x^{2} + 4

Expand

2 x (x + 2) + 2 x (x + 1) : 4 x^{2} + 6 x

4 x^{2} + 6 x = x^{2} + 4

Move

4

to the left side

4 x^{2} + 6 x - 4 = x^{2}

Move

x^{2}

to the left side

3 x^{2} + 6 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 4) = 221

x_{1, 2} = \frac{- 6 \pm 2 21}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 2 21}{2 \cdot 3}, x_{2} = \frac{- 6 - 2 21}{2 \cdot 3}

x = \frac{- 6 + 2 21}{2 \cdot 3} : \frac{- 3 + 21}{3}

x = \frac{- 6 - 2 21}{2 \cdot 3} : - \frac{3 + 21}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 21}{3}, x = - \frac{3 + 21}{3}

\frac{x ^{2}}{4} - \frac{3 x}{2} + 2 = 0

x^{2} + 7 x - 24 = 5 x

7 x^{2} + 3 x - 9 = 0

2 x^{2} + 1 x - 2 = 0

z^{2} - 14 z + 48 = 0

What are the solutions to the equation 2x(x+2)+2x(x+1)=x^2+4 ?
The solutions to the equation 2x(x+2)+2x(x+1)=x^2+4 are x=(-3+sqrt(21))/3 ,x=-(3+sqrt(21))/3
Find the zeros of 2x(x+2)+2x(x+1)=x^2+4
The zeros of 2x(x+2)+2x(x+1)=x^2+4 are x=(-3+sqrt(21))/3 ,x=-(3+sqrt(21))/3