What are the solutions to the equation 5x-1=x^2+2x+1 ?

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

Find the zeros of 5x-1=x^2+2x+1

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

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5 x - 1 = x^{2} + 2 x + 1

x = 2, x = 1

Solution steps

5 x - 1 = x^{2} + 2 x + 1

Switch sides

x^{2} + 2 x + 1 = 5 x - 1

Move

1

to the left side

x^{2} + 2 x + 2 = 5 x

Move

5 x

to the left side

x^{2} - 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

x_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

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

x = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 2, x = 1

x^{2} + 9 x - 10 = 0

15 (- 3 x + 2)^{2} + 4 (- 3 x + 2) - 3 = 0

2 x^{2} = 8 x - 5

2 x^{2} = 8 x - 8

- 2 = 1.1 t + (0.5) (- 9.8) t^{2}

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