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

The solutions to the equation x(x+1)=4-(x+2)^2 are x=0,x=-5/2

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

The zeros of x(x+1)=4-(x+2)^2 are x=0,x=-5/2

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

x = 0, x = - \frac{5}{2}

Decimal Notation

x = 0, x = - 2.5

Solution steps

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

Expand

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

Expand

4 - (x + 2)^{2} : - x^{2} - 4 x

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

Move

4 x

to the left side

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

Move

x^{2}

to the left side

2 x^{2} + 5 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 0 = 5

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

Separate the solutions

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

x = \frac{- 5 + 5}{2 \cdot 2} : 0

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

The solutions to the quadratic equation are:

x = 0, x = - \frac{5}{2}

3 x^{2} + 24 x + 32 = 0

x, 2 x^{2} y = 5

x, 3 x^{2} = 225

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

x^{2} + 4 x + 7 = 0

What are the solutions to the equation x(x+1)=4-(x+2)^2 ?
The solutions to the equation x(x+1)=4-(x+2)^2 are x=0,x=-5/2
Find the zeros of x(x+1)=4-(x+2)^2
The zeros of x(x+1)=4-(x+2)^2 are x=0,x=-5/2