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

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

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

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

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

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

Solution steps

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

Switch sides

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

∣ 1 - 2 t ∣ + 4 = 11

8 x^{2} - 14 = - 11

\frac{( 2 x - 3 )}{( x + 1 )} \leq 1

- \frac{3}{4} + \frac{7}{2} = - 43 + 27

(- 1)^{0}

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