What are the solutions to the equation 16x^2-12x+4=0 ?

The solutions to the equation 16x^2-12x+4=0 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

Find the zeros of 16x^2-12x+4=0

The zeros of 16x^2-12x+4=0 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

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16 x^{2} - 12 x + 4 = 0

x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

Solution steps

16 x^{2} - 12 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 16 \cdot 4}{2 \cdot 16}

Simplify

(- 12)^{2} - 4 \cdot 16 \cdot 4 : 47 i

x_{1, 2} = \frac{- ( - 12 ) \pm 4 7 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 7 i}{2 \cdot 16}, x_{2} = \frac{- ( - 12 ) - 4 7 i}{2 \cdot 16}

x = \frac{- ( - 12 ) + 4 7 i}{2 \cdot 16} : \frac{3}{8} + i \frac{7}{8}

x = \frac{- ( - 12 ) - 4 7 i}{2 \cdot 16} : \frac{3}{8} - i \frac{7}{8}

The solutions to the quadratic equation are:

x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

\frac{x ^{2} - 4}{x ^{2} - 36} \leq 0

\frac{- 3 m}{( 4 m - 7 ) ( 2 m - 5 )} - \frac{8}{( 6 m + 1 ) ( 2 m - 5 )}

5 x - 9 \leq 2 x + 18

\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}

∣ x + 5 ∣ > 2

What are the solutions to the equation 16x^2-12x+4=0 ?
The solutions to the equation 16x^2-12x+4=0 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8
Find the zeros of 16x^2-12x+4=0
The zeros of 16x^2-12x+4=0 are x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8