What are the solutions to the equation r^2-4r+3=0 ?

The solutions to the equation r^2-4r+3=0 are r=3,r=1

Find the zeros of r^2-4r+3=0

The zeros of r^2-4r+3=0 are r=3,r=1

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r^{2} - 4 r + 3 = 0

r = 3, r = 1

Solution steps

r^{2} - 4 r + 3 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

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

r = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

r = 3, r = 1

2 (\frac{1}{5} m - \frac{2}{5}) + \frac{3}{5}

108 \geq - 6 + 2 (1 - 7 x)

i^{- 27}

(3)^{- 2}

36 x^{2} + 0.5 + \frac{1}{576 x ^{2}}

What are the solutions to the equation r^2-4r+3=0 ?
The solutions to the equation r^2-4r+3=0 are r=3,r=1
Find the zeros of r^2-4r+3=0
The zeros of r^2-4r+3=0 are r=3,r=1