What are the solutions to the equation x^2+2x+1=16 ?

The solutions to the equation x^2+2x+1=16 are x=3,x=-5

Find the zeros of x^2+2x+1=16

The zeros of x^2+2x+1=16 are x=3,x=-5

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x^{2} + 2 x + 1 = 16

x = 3, x = - 5

Solution steps

x^{2} + 2 x + 1 = 16

Move

16

to the left side

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

\frac{2}{3} y + 15 = 9

- 7 - 6 x < - 31

- 1 + \frac{5}{- 1}

∣ x + 2 ∣ + 9 = 4

2^{- 7} \cdot 2^{12}

What are the solutions to the equation x^2+2x+1=16 ?
The solutions to the equation x^2+2x+1=16 are x=3,x=-5
Find the zeros of x^2+2x+1=16
The zeros of x^2+2x+1=16 are x=3,x=-5