What are the solutions to the equation t^2-2t+8=0 ?

The solutions to the equation t^2-2t+8=0 are t=1+sqrt(7)i,t=1-sqrt(7)i

Find the zeros of t^2-2t+8=0

The zeros of t^2-2t+8=0 are t=1+sqrt(7)i,t=1-sqrt(7)i

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t^{2} - 2 t + 8 = 0

t = 1 + 7 i, t = 1 - 7 i

Solution steps

t^{2} - 2 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 8 : 27 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 1 + 7 i, t = 1 - 7 i

- 31 = 2 - 3 a^{2}

x, 3 x^{2} - 2 x + 2 = 0

x, f^{- 1} (x) = x^{2} + 5

(x - 1) (2 x - 3) = 1

84 = \frac{1}{2} x (3 x + 3)

What are the solutions to the equation t^2-2t+8=0 ?
The solutions to the equation t^2-2t+8=0 are t=1+sqrt(7)i,t=1-sqrt(7)i
Find the zeros of t^2-2t+8=0
The zeros of t^2-2t+8=0 are t=1+sqrt(7)i,t=1-sqrt(7)i