What are the solutions to the equation x^2+1-2x=0 ?

The solutions to the equation x^2+1-2x=0 are x=1

Find the zeros of x^2+1-2x=0

The zeros of x^2+1-2x=0 are x=1

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x^{2} + 1 - 2 x = 0

x = 1

Solution steps

x^{2} + 1 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 2 x + 1 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

(x^{0.5})^{2}

- 12 (x + 6.2) = 60

\frac{3 x ^{12} y ^{10}}{5 x ^{6} y ^{3}}

\frac{2}{4} \cdot \frac{2}{6}

60 \cdot \frac{π}{180}

What are the solutions to the equation x^2+1-2x=0 ?
The solutions to the equation x^2+1-2x=0 are x=1
Find the zeros of x^2+1-2x=0
The zeros of x^2+1-2x=0 are x=1