What are the solutions to the equation 16x^2-32x+17=0 ?

The solutions to the equation 16x^2-32x+17=0 are x=1+i 1/4 ,x=1-i 1/4

Find the zeros of 16x^2-32x+17=0

The zeros of 16x^2-32x+17=0 are x=1+i 1/4 ,x=1-i 1/4

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16 x^{2} - 32 x + 17 = 0

x = 1 + i \frac{1}{4}, x = 1 - i \frac{1}{4}

Solution steps

16 x^{2} - 32 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 16 \cdot 17}{2 \cdot 16}

Simplify

(- 32)^{2} - 4 \cdot 16 \cdot 17 : 8 i

x_{1, 2} = \frac{- ( - 32 ) \pm 8 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 8 i}{2 \cdot 16}, x_{2} = \frac{- ( - 32 ) - 8 i}{2 \cdot 16}

x = \frac{- ( - 32 ) + 8 i}{2 \cdot 16} : 1 + i \frac{1}{4}

x = \frac{- ( - 32 ) - 8 i}{2 \cdot 16} : 1 - i \frac{1}{4}

The solutions to the quadratic equation are:

x = 1 + i \frac{1}{4}, x = 1 - i \frac{1}{4}

x^{2} - 5 x = - 8

x, x^{2} + x y + y^{2} = 7

x^{2} - x

x^{2} + 10 = 11 x

x, z^{2} = x^{2} + y^{2}

What are the solutions to the equation 16x^2-32x+17=0 ?
The solutions to the equation 16x^2-32x+17=0 are x=1+i 1/4 ,x=1-i 1/4
Find the zeros of 16x^2-32x+17=0
The zeros of 16x^2-32x+17=0 are x=1+i 1/4 ,x=1-i 1/4