What are the solutions to the equation x^2-4x+1=6 ?

The solutions to the equation x^2-4x+1=6 are x=5,x=-1

Find the zeros of x^2-4x+1=6

The zeros of x^2-4x+1=6 are x=5,x=-1

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x^{2} - 4 x + 1 = 6

x = 5, x = - 1

Solution steps

x^{2} - 4 x + 1 = 6

Move

6

to the left side

x^{2} - 4 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

x = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

x = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 5, x = - 1

5 x - 3 \leq 7 - 3 x

x^{3} y^{3} (3 y^{3})

2 < 2 x + 2 \leq 8

\frac{7 k ^{- 8} \cdot 3 k ^{- 2}}{6 k ^{2}}

- 9 k^{2} - 10 k + 21 = 0

What are the solutions to the equation x^2-4x+1=6 ?
The solutions to the equation x^2-4x+1=6 are x=5,x=-1
Find the zeros of x^2-4x+1=6
The zeros of x^2-4x+1=6 are x=5,x=-1