What are the solutions to the equation 16t^2-4t+5=0 ?

The solutions to the equation 16t^2-4t+5=0 are t= 1/8+i(sqrt(19))/8 ,t= 1/8-i(sqrt(19))/8

Find the zeros of 16t^2-4t+5=0

The zeros of 16t^2-4t+5=0 are t= 1/8+i(sqrt(19))/8 ,t= 1/8-i(sqrt(19))/8

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16 t^{2} - 4 t + 5 = 0

t = \frac{1}{8} + i \frac{19}{8}, t = \frac{1}{8} - i \frac{19}{8}

Solution steps

16 t^{2} - 4 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 16 \cdot 5}{2 \cdot 16}

Simplify

(- 4)^{2} - 4 \cdot 16 \cdot 5 : 419 i

t_{1, 2} = \frac{- ( - 4 ) \pm 4 19 i}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 4 19 i}{2 \cdot 16}, t_{2} = \frac{- ( - 4 ) - 4 19 i}{2 \cdot 16}

t = \frac{- ( - 4 ) + 4 19 i}{2 \cdot 16} : \frac{1}{8} + i \frac{19}{8}

t = \frac{- ( - 4 ) - 4 19 i}{2 \cdot 16} : \frac{1}{8} - i \frac{19}{8}

The solutions to the quadratic equation are:

t = \frac{1}{8} + i \frac{19}{8}, t = \frac{1}{8} - i \frac{19}{8}

x^{2} - 6 x + 10 = 5

(- 2) \frac{2 + 3 i}{1 + i} + (2 - i)^{2}

2 x + 1 < 1

3 x^{2} + x - 14 = 0

2 x^{2} - x = - 1

What are the solutions to the equation 16t^2-4t+5=0 ?
The solutions to the equation 16t^2-4t+5=0 are t= 1/8+i(sqrt(19))/8 ,t= 1/8-i(sqrt(19))/8
Find the zeros of 16t^2-4t+5=0
The zeros of 16t^2-4t+5=0 are t= 1/8+i(sqrt(19))/8 ,t= 1/8-i(sqrt(19))/8