What are the solutions to the equation x^2-2x+5=0 ?

The solutions to the equation x^2-2x+5=0 are x=1+2i,x=1-2i

Find the zeros of x^2-2x+5=0

The zeros of x^2-2x+5=0 are x=1+2i,x=1-2i

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

x = 1 + 2 i, x = 1 - 2 i

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 2 i, x = 1 - 2 i

x^{2} - 17 x + 30

w^{2} + 3 w - 28 = 0

2 r^{2} - r - 10 = 0

\frac{68}{( 1 - i ) ( 2 - i ) ( 3 - i )}

\frac{48}{2}

What are the solutions to the equation x^2-2x+5=0 ?
The solutions to the equation x^2-2x+5=0 are x=1+2i,x=1-2i
Find the zeros of x^2-2x+5=0
The zeros of x^2-2x+5=0 are x=1+2i,x=1-2i