What are the solutions to the equation x(x+2)=-1+3(x+2) ?

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

Find the zeros of x(x+2)=-1+3(x+2)

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

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x (x + 2) = - 1 + 3 (x + 2)

x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

Solution steps

x (x + 2) = - 1 + 3 (x + 2)

Expand

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

Expand

- 1 + 3 (x + 2) : 3 x + 5

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

Move

5

to the left side

x^{2} + 2 x - 5 = 3 x

Move

3 x

to the left side

x^{2} - x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 5) = 21

x_{1, 2} = \frac{- ( - 1 ) \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 21}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 21}{2 \cdot 1}

x = \frac{- ( - 1 ) + 21}{2 \cdot 1} : \frac{1 + 21}{2}

x = \frac{- ( - 1 ) - 21}{2 \cdot 1} : \frac{1 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

x (x - 3) (x + 2) \leq 0

2 - 9

∣ x - 6 ∣ = 3

x^{2} - 15 x + 56 = 0

- \frac{4}{3} + 5

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