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

The solutions to the equation t^2-t+2=0 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2

Find the zeros of t^2-t+2=0

The zeros of t^2-t+2=0 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2

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t^{2} - t + 2 = 0

t = \frac{1}{2} + i \frac{7}{2}, t = \frac{1}{2} - i \frac{7}{2}

Solution steps

t^{2} - t + 2 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

t_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, t_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

t = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

t = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

The solutions to the quadratic equation are:

t = \frac{1}{2} + i \frac{7}{2}, t = \frac{1}{2} - i \frac{7}{2}

3 x^{2} - 5 x - 8 = 0

2000 (1 + \frac{0.09}{12})^{60}

\frac{x ^{2}}{4} - \frac{5 x}{2} + 4 = 0

35 k^{2} - 22 k + 7 = 4

\frac{x + 2}{2} + 2 = \frac{x + 3}{5}

What are the solutions to the equation t^2-t+2=0 ?
The solutions to the equation t^2-t+2=0 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2
Find the zeros of t^2-t+2=0
The zeros of t^2-t+2=0 are t= 1/2+i(sqrt(7))/2 ,t= 1/2-i(sqrt(7))/2