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

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

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

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

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

x = 1 + 4 i, x = 1 - 4 i

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 17 : 8 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 4 i, x = 1 - 4 i

(x - 4) (x + 5) < (x - 3) (x - 2)

3 (1 - 3 x) = 2 (- 4 x + 7)

27 + 3 - \frac{45}{5 + 16}

\frac{π}{3} + \frac{π}{2}

a^{2} + 6 a - 7 = 0

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