What are the solutions to the equation x^2+2x+1=17 ?

The solutions to the equation x^2+2x+1=17 are x=-1+sqrt(17),x=-1-sqrt(17)

Find the zeros of x^2+2x+1=17

The zeros of x^2+2x+1=17 are x=-1+sqrt(17),x=-1-sqrt(17)

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x^{2} + 2 x + 1 = 17

x = - 1 + 17, x = - 1 - 17

Decimal Notation

x = 3.12310 \dots, x = - 5.12310 \dots

Solution steps

x^{2} + 2 x + 1 = 17

Move

17

to the left side

x^{2} + 2 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 16 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 16) = 217

x_{1, 2} = \frac{- 2 \pm 2 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 17}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 17}{2 \cdot 1}

x = \frac{- 2 + 2 17}{2 \cdot 1} : - 1 + 17

x = \frac{- 2 - 2 17}{2 \cdot 1} : - 1 - 17

The solutions to the quadratic equation are:

x = - 1 + 17, x = - 1 - 17

2 ∣ x - 3 ∣ - 3 > 15

x^{2} + 4 x + 10 = 0

\frac{x y ^{9}}{3 y ^{- 2}} \cdot \frac{- 7 y}{21 x ^{5}}

4 (6 + 2 x) + 3 \geq 9 - x

(\frac{2}{5} m^{4} n + \frac{1}{3} mn) (\frac{3}{2} m^{2} n + 2 mn - 3)

What are the solutions to the equation x^2+2x+1=17 ?
The solutions to the equation x^2+2x+1=17 are x=-1+sqrt(17),x=-1-sqrt(17)
Find the zeros of x^2+2x+1=17
The zeros of x^2+2x+1=17 are x=-1+sqrt(17),x=-1-sqrt(17)