What are the solutions to the equation 4t^2-4t+17=0 ?

The solutions to the equation 4t^2-4t+17=0 are t= 1/2+2i,t= 1/2-2i

Find the zeros of 4t^2-4t+17=0

The zeros of 4t^2-4t+17=0 are t= 1/2+2i,t= 1/2-2i

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4 t^{2} - 4 t + 17 = 0

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

Solution steps

4 t^{2} - 4 t + 17 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 17}{2 \cdot 4}

Simplify

(- 4)^{2} - 4 \cdot 4 \cdot 17 : 16 i

t_{1, 2} = \frac{- ( - 4 ) \pm 16 i}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 16 i}{2 \cdot 4}, t_{2} = \frac{- ( - 4 ) - 16 i}{2 \cdot 4}

t = \frac{- ( - 4 ) + 16 i}{2 \cdot 4} : \frac{1}{2} + 2 i

t = \frac{- ( - 4 ) - 16 i}{2 \cdot 4} : \frac{1}{2} - 2 i

The solutions to the quadratic equation are:

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

- 16 t^{2} + 12 t + 10 = 0

5 (m - 1) - 7 m (m - 1)

49 x^{4} - 14 x^{3} - 28 x

- k + 3 k

(t - 6) (t + 6) = (t - 2) (t - 4)

What are the solutions to the equation 4t^2-4t+17=0 ?
The solutions to the equation 4t^2-4t+17=0 are t= 1/2+2i,t= 1/2-2i
Find the zeros of 4t^2-4t+17=0
The zeros of 4t^2-4t+17=0 are t= 1/2+2i,t= 1/2-2i