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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x^{2} + x - 1 = 0

y^{2} + 2 y = 3

x^{2} + 5 x + 6 = 0

y^{2} + 49 = 0

x^{2} + 5 x + 4 = 0

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