What are the solutions to the equation x^2+3=x ?

The solutions to the equation x^2+3=x are x= 1/2+i(sqrt(11))/2 ,x= 1/2-i(sqrt(11))/2

Find the zeros of x^2+3=x

The zeros of x^2+3=x are x= 1/2+i(sqrt(11))/2 ,x= 1/2-i(sqrt(11))/2

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x^{2} + 3 = x

x = \frac{1}{2} + i \frac{11}{2}, x = \frac{1}{2} - i \frac{11}{2}

Solution steps

x^{2} + 3 = x

Move

x

to the left side

x^{2} + 3 - x = 0

Write in the standard form

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

x^{2} - x + 3 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 3 : 11 i

x_{1, 2} = \frac{- ( - 1 ) \pm 11 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 11 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 11 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 11 i}{2 \cdot 1} : \frac{1}{2} + i \frac{11}{2}

x = \frac{- ( - 1 ) - 11 i}{2 \cdot 1} : \frac{1}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{11}{2}, x = \frac{1}{2} - i \frac{11}{2}

- \frac{4}{7} + y = - \frac{2}{3}

(\frac{1}{3})^{- \frac{2}{5}} \cdot 9 6^{- \frac{2}{5}}

\frac{- 4}{- 5 - 4 i}

4 - u^{2}

\frac{6}{7} - \frac{1}{3}

What are the solutions to the equation x^2+3=x ?
The solutions to the equation x^2+3=x are x= 1/2+i(sqrt(11))/2 ,x= 1/2-i(sqrt(11))/2
Find the zeros of x^2+3=x
The zeros of x^2+3=x are x= 1/2+i(sqrt(11))/2 ,x= 1/2-i(sqrt(11))/2