What are the solutions to the equation x+(x+1)+(xx+1+1)=27 ?

The solutions to the equation x+(x+1)+(xx+1+1)=27 are x=4,x=-6

Find the zeros of x+(x+1)+(xx+1+1)=27

The zeros of x+(x+1)+(xx+1+1)=27 are x=4,x=-6

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x + (x + 1) + (xx + 1 + 1) = 27

x = 4, x = - 6

Solution steps

x + (x + 1) + (xx + 1 + 1) = 27

Expand

x + (x + 1) + (xx + 1 + 1) : x^{2} + 2 x + 3

x^{2} + 2 x + 3 = 27

Move

27

to the left side

x^{2} + 2 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 1}, x_{2} = \frac{- 2 - 10}{2 \cdot 1}

x = \frac{- 2 + 10}{2 \cdot 1} : 4

x = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

\frac{3}{16}

\frac{1}{2} \cdot \frac{5}{6}

\frac{3 ^{8}}{3 ^{3}}

15 t^{2} - t - 6 = 0

2 - \frac{2}{3} x = 4 + \frac{7}{8} x

What are the solutions to the equation x+(x+1)+(xx+1+1)=27 ?
The solutions to the equation x+(x+1)+(xx+1+1)=27 are x=4,x=-6
Find the zeros of x+(x+1)+(xx+1+1)=27
The zeros of x+(x+1)+(xx+1+1)=27 are x=4,x=-6