What are the solutions to the equation x(x+1)=12 ?

The solutions to the equation x(x+1)=12 are x=3,x=-4

Find the zeros of x(x+1)=12

The zeros of x(x+1)=12 are x=3,x=-4

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x (x + 1) = 12

x = 3, x = - 4

Solution steps

x (x + 1) = 12

Expand

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

x^{2} + x = 12

Move

12

to the left side

x^{2} + x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 12) = 7

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

Separate the solutions

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

x = \frac{- 1 + 7}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

x = 3, x = - 4

y^{2} - 12 y + 32

(\frac{x}{y})^{5}

(1 - 3) (1 + 3)

9 a^{5} + 27 a^{4} + 63 a^{2}

x - 4 = 1

What are the solutions to the equation x(x+1)=12 ?
The solutions to the equation x(x+1)=12 are x=3,x=-4
Find the zeros of x(x+1)=12
The zeros of x(x+1)=12 are x=3,x=-4