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

The solutions to the equation x(x+1)=2 are x=1,x=-2

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

The zeros of x(x+1)=2 are x=1,x=-2

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x (x + 1) = 2

x = 1, x = - 2

Solution steps

x (x + 1) = 2

Expand

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

x^{2} + x = 2

Move

2

to the left side

x^{2} + x - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 1 + 3}{2 \cdot 1} : 1

x = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

5 x^{2} + 6 x - 63 = 0

u^{2} = 4

w^{2} - 17 w + 72

x^{3} + 12 x^{2} + 9 x - 22

(x^{4} y^{2})^{\frac{1}{2}}

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