What are the solutions to the equation t^2-4t+8=0 ?

The solutions to the equation t^2-4t+8=0 are t=2+2i,t=2-2i

Find the zeros of t^2-4t+8=0

The zeros of t^2-4t+8=0 are t=2+2i,t=2-2i

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t^{2} - 4 t + 8 = 0

t = 2 + 2 i, t = 2 - 2 i

Solution steps

t^{2} - 4 t + 8 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 4 i}{2 \cdot 1}, t_{2} = \frac{- ( - 4 ) - 4 i}{2 \cdot 1}

t = \frac{- ( - 4 ) + 4 i}{2 \cdot 1} : 2 + 2 i

t = \frac{- ( - 4 ) - 4 i}{2 \cdot 1} : 2 - 2 i

The solutions to the quadratic equation are:

t = 2 + 2 i, t = 2 - 2 i

b^{2} - 7 b + 6

8^{- 2} = \frac{1}{64}

(- 4 x y)^{3} (- 2 x^{2})^{3}

13 x - 5 (x + 2) = 4 (2 x - 1) + 7

\frac{1}{x - 1} \leq 1

What are the solutions to the equation t^2-4t+8=0 ?
The solutions to the equation t^2-4t+8=0 are t=2+2i,t=2-2i
Find the zeros of t^2-4t+8=0
The zeros of t^2-4t+8=0 are t=2+2i,t=2-2i