What are the solutions to the equation x=x^2-6x+5 ?

The solutions to the equation x=x^2-6x+5 are x=(7+sqrt(29))/2 ,x=(7-sqrt(29))/2

Find the zeros of x=x^2-6x+5

The zeros of x=x^2-6x+5 are x=(7+sqrt(29))/2 ,x=(7-sqrt(29))/2

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x = x^{2} - 6 x + 5

x = \frac{7 + 29}{2}, x = \frac{7 - 29}{2}

Solution steps

x = x^{2} - 6 x + 5

Switch sides

x^{2} - 6 x + 5 = x

Move

x

to the left side

x^{2} - 7 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 5 = 29

x_{1, 2} = \frac{- ( - 7 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 29}{2 \cdot 1}

x = \frac{- ( - 7 ) + 29}{2 \cdot 1} : \frac{7 + 29}{2}

x = \frac{- ( - 7 ) - 29}{2 \cdot 1} : \frac{7 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 29}{2}, x = \frac{7 - 29}{2}

198

\frac{2}{3} x - \frac{3}{4} + \frac{1}{4} x = x - \frac{5}{4}

3 n^{2} + 4 n + 1

- 3 x - 6 < 12

- 14 < u - 15

What are the solutions to the equation x=x^2-6x+5 ?
The solutions to the equation x=x^2-6x+5 are x=(7+sqrt(29))/2 ,x=(7-sqrt(29))/2
Find the zeros of x=x^2-6x+5
The zeros of x=x^2-6x+5 are x=(7+sqrt(29))/2 ,x=(7-sqrt(29))/2