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

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

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

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

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x^{2} - x = x - 5

x = 1 + 2 i, x = 1 - 2 i

Solution steps

x^{2} - x = x - 5

Move

5

to the left side

x^{2} - x + 5 = x

Move

x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 2 i, x = 1 - 2 i

x - 4 = \frac{( x - 22 ) ^{2}}{3}

12 x^{2} + 16 = 0

x^{2} - 24 x - 82 = 0

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

8^{2} + x^{2} = 1 3^{2}

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