What are the solutions to the equation x(1+x)=6 ?

The solutions to the equation x(1+x)=6 are x=2,x=-3

Find the zeros of x(1+x)=6

The zeros of x(1+x)=6 are x=2,x=-3

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x (1 + x) = 6

x = 2, x = - 3

Solution steps

x (1 + x) = 6

Expand

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

x + x^{2} = 6

Move

6

to the left side

x + x^{2} - 6 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + x - 6 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

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

Separate the solutions

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

x = \frac{- 1 + 5}{2 \cdot 1} : 2

x = \frac{- 1 - 5}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 2, x = - 3

∣ x - 2 ∣ = 3 x + 1

\frac{5 ^{6}}{5}

\frac{6 4 ^{\frac{7}{10}}}{6 4 ^{\frac{1}{5}}}

\frac{1}{3} x - \frac{5}{6} = - \frac{2}{3}

e^{- 2 l n (4)}

What are the solutions to the equation x(1+x)=6 ?
The solutions to the equation x(1+x)=6 are x=2,x=-3
Find the zeros of x(1+x)=6
The zeros of x(1+x)=6 are x=2,x=-3