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

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

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

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

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

y = 2, y = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2, y = 1

(x - 2)^{2} = 1

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- 5 (2 x - 4) + 2 \geq - 13

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