What are the solutions to the equation 16t^2-4t+3=0 ?

The solutions to the equation 16t^2-4t+3=0 are t= 1/8+i(sqrt(11))/8 ,t= 1/8-i(sqrt(11))/8

Find the zeros of 16t^2-4t+3=0

The zeros of 16t^2-4t+3=0 are t= 1/8+i(sqrt(11))/8 ,t= 1/8-i(sqrt(11))/8

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16 t^{2} - 4 t + 3 = 0

t = \frac{1}{8} + i \frac{11}{8}, t = \frac{1}{8} - i \frac{11}{8}

Solution steps

16 t^{2} - 4 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 16 \cdot 3}{2 \cdot 16}

Simplify

(- 4)^{2} - 4 \cdot 16 \cdot 3 : 411 i

t_{1, 2} = \frac{- ( - 4 ) \pm 4 11 i}{2 \cdot 16}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 4 11 i}{2 \cdot 16}, t_{2} = \frac{- ( - 4 ) - 4 11 i}{2 \cdot 16}

t = \frac{- ( - 4 ) + 4 11 i}{2 \cdot 16} : \frac{1}{8} + i \frac{11}{8}

t = \frac{- ( - 4 ) - 4 11 i}{2 \cdot 16} : \frac{1}{8} - i \frac{11}{8}

The solutions to the quadratic equation are:

t = \frac{1}{8} + i \frac{11}{8}, t = \frac{1}{8} - i \frac{11}{8}

n - \frac{11}{12} \geq - \frac{5}{18}

v - 5 + 3 (v + 4) = 7 (v - 8)

\frac{6 - 5 x}{3 + x} \leq \frac{1}{2}

12 - x^{2} + x = 0

- 5 \cdot 6

What are the solutions to the equation 16t^2-4t+3=0 ?
The solutions to the equation 16t^2-4t+3=0 are t= 1/8+i(sqrt(11))/8 ,t= 1/8-i(sqrt(11))/8
Find the zeros of 16t^2-4t+3=0
The zeros of 16t^2-4t+3=0 are t= 1/8+i(sqrt(11))/8 ,t= 1/8-i(sqrt(11))/8