What are the solutions to the equation x^2+2x+1=49 ?

The solutions to the equation x^2+2x+1=49 are x=6,x=-8

Find the zeros of x^2+2x+1=49

The zeros of x^2+2x+1=49 are x=6,x=-8

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x^{2} + 2 x + 1 = 49

x = 6, x = - 8

Solution steps

x^{2} + 2 x + 1 = 49

Move

49

to the left side

x^{2} + 2 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 48) = 14

x_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 14}{2 \cdot 1}, x_{2} = \frac{- 2 - 14}{2 \cdot 1}

x = \frac{- 2 + 14}{2 \cdot 1} : 6

x = \frac{- 2 - 14}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

x^{2} - 7 x - 12 = 0

2 - \frac{3}{4}

\frac{d}{4} = 10

3 c + 9 5 d

\frac{x + 1}{2} + \frac{5 + x}{6} = 1 + \frac{9 - 2 x}{3}

What are the solutions to the equation x^2+2x+1=49 ?
The solutions to the equation x^2+2x+1=49 are x=6,x=-8
Find the zeros of x^2+2x+1=49
The zeros of x^2+2x+1=49 are x=6,x=-8