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

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

Find the zeros of t^2-t+1=1

The zeros of t^2-t+1=1 are t=1,t=0

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t^{2} - t + 1 = 1

t = 1, t = 0

Solution steps

t^{2} - t + 1 = 1

Move

1

to the left side

t^{2} - t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

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

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

Separate the solutions

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

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

t = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

t = 1, t = 0

3 (2 - 2 x) = (x - 4) (x - 2)

x^{2} + 14 x + 40 = 0

101^{2} + x^{2} = 1 4^{2}

4 x^{2} + 77 = - 36

w, w^{2} + 4 w = 0

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