What are the solutions to the equation 0=x^2-4x+5 ?

The solutions to the equation 0=x^2-4x+5 are x=2+i,x=2-i

Find the zeros of 0=x^2-4x+5

The zeros of 0=x^2-4x+5 are x=2+i,x=2-i

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0 = x^{2} - 4 x + 5

x = 2 + i, x = 2 - i

Solution steps

0 = x^{2} - 4 x + 5

Switch sides

x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 + i, x = 2 - i

t, h = \frac{1}{2} g t^{2} + v t

x^{2} + 4 x + 4 = 0

2 y^{2} = 32

9 (2 x - 3)^{2} = 16

x^{2} - 19 x = 9.75

What are the solutions to the equation 0=x^2-4x+5 ?
The solutions to the equation 0=x^2-4x+5 are x=2+i,x=2-i
Find the zeros of 0=x^2-4x+5
The zeros of 0=x^2-4x+5 are x=2+i,x=2-i