What are the solutions to the equation x^2-x+1=3 ?

The solutions to the equation x^2-x+1=3 are x=2,x=-1

Find the zeros of x^2-x+1=3

The zeros of x^2-x+1=3 are x=2,x=-1

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x^{2} - x + 1 = 3

x = 2, x = - 1

Solution steps

x^{2} - x + 1 = 3

Move

3

to the left side

x^{2} - x - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

x = \frac{- ( - 1 ) + 3}{2 \cdot 1} : 2

x = \frac{- ( - 1 ) - 3}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

\frac{1}{3} (2 y + 1) + \frac{1}{2} y = \frac{2}{5} (1 - 2 y) - 4

s^{3} + 1

- 220 + 218 - 25

\frac{5}{6} + 1

\frac{7}{2} = \frac{t + 5}{10}

What are the solutions to the equation x^2-x+1=3 ?
The solutions to the equation x^2-x+1=3 are x=2,x=-1
Find the zeros of x^2-x+1=3
The zeros of x^2-x+1=3 are x=2,x=-1