What are the solutions to the equation 16x^2-4x+14=0 ?

The solutions to the equation 16x^2-4x+14=0 are x= 1/8+i(sqrt(55))/8 ,x= 1/8-i(sqrt(55))/8

Find the zeros of 16x^2-4x+14=0

The zeros of 16x^2-4x+14=0 are x= 1/8+i(sqrt(55))/8 ,x= 1/8-i(sqrt(55))/8

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16 x^{2} - 4 x + 14 = 0

x = \frac{1}{8} + i \frac{55}{8}, x = \frac{1}{8} - i \frac{55}{8}

Solution steps

16 x^{2} - 4 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 16 \cdot 14}{2 \cdot 16}

Simplify

(- 4)^{2} - 4 \cdot 16 \cdot 14 : 455 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 55 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 55 i}{2 \cdot 16}, x_{2} = \frac{- ( - 4 ) - 4 55 i}{2 \cdot 16}

x = \frac{- ( - 4 ) + 4 55 i}{2 \cdot 16} : \frac{1}{8} + i \frac{55}{8}

x = \frac{- ( - 4 ) - 4 55 i}{2 \cdot 16} : \frac{1}{8} - i \frac{55}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8} + i \frac{55}{8}, x = \frac{1}{8} - i \frac{55}{8}

0 \cdot \frac{1}{3}

x^{2} + 9 < 6 x

- x + 3 = - 5

64 x \leq x^{3}

- \frac{6 ( - 1 ) + 4}{( 3 ( - 1 ) ^{2} + 4 ( - 1 ) ) ^{2}}

What are the solutions to the equation 16x^2-4x+14=0 ?
The solutions to the equation 16x^2-4x+14=0 are x= 1/8+i(sqrt(55))/8 ,x= 1/8-i(sqrt(55))/8
Find the zeros of 16x^2-4x+14=0
The zeros of 16x^2-4x+14=0 are x= 1/8+i(sqrt(55))/8 ,x= 1/8-i(sqrt(55))/8