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

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

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

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

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

r = 2, r = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

r = 2, r = 1

12 (\frac{π}{3})

\frac{x + x ^{- 1}}{x ^{- 1} - y}

3 v + 5 > 20

\frac{2 x - 1}{x - 3} \geq 1

m^{2} + 3 m - 10

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