What are the solutions to the equation solvefor x,150=x^2+2x+1 ?

The solutions to the equation solvefor x,150=x^2+2x+1 are x=-1+5sqrt(6),x=-1-5sqrt(6)

Find the zeros of solvefor x,150=x^2+2x+1

The zeros of solvefor x,150=x^2+2x+1 are x=-1+5sqrt(6),x=-1-5sqrt(6)

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x, 150 = x^{2} + 2 x + 1

x = - 1 + 56, x = - 1 - 56

Decimal Notation

x = 11.24744 \dots, x = - 13.24744 \dots

Solution steps

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

Switch sides

x^{2} + 2 x + 1 = 150

Move

150

to the left side

x^{2} + 2 x - 149 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 149 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 149) = 106

x_{1, 2} = \frac{- 2 \pm 10 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 10 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 10 6}{2 \cdot 1}

x = \frac{- 2 + 10 6}{2 \cdot 1} : - 1 + 56

x = \frac{- 2 - 10 6}{2 \cdot 1} : - 1 - 56

The solutions to the quadratic equation are:

x = - 1 + 56, x = - 1 - 56

- (x - 2)^{2} + 4 = - 5

4 x^{2} - 12 x + 16 = 0

x^{2} = 6 x - 7

y, z = 1 - y^{2}

7 x^{2} - 64 = 0

What are the solutions to the equation solvefor x,150=x^2+2x+1 ?
The solutions to the equation solvefor x,150=x^2+2x+1 are x=-1+5sqrt(6),x=-1-5sqrt(6)
Find the zeros of solvefor x,150=x^2+2x+1
The zeros of solvefor x,150=x^2+2x+1 are x=-1+5sqrt(6),x=-1-5sqrt(6)