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

Refine

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

6 r^{2} - 11 r p + 5 p^{2}

4 x - 4 + x = 5 x - 4

2.38 a - (3.5 + 1.38 a) = 0.27

3 x^{2} - 5 x - 12 = 0

4 x + 188 = 6 x + 216

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