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

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

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

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

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

r = 2 + i, r = 2 - i

Solution steps

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

Solve with the quadratic formula

r_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 2 + i, r = 2 - i

- 2 ∣ x + 1 ∣ + 5 = 1

9^{- 2}

25 x - 4 = 35 x + 2

x^{2} = - 15 x - 56

3 a^{3} - 1 - a^{2} + 3 a

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