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

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

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

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

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

y = 3, y = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 3, y = 1

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

12 m + 18 m^{2}

∣ 5 x - 3 ∣ < 7

8 \cdot 3^{- 2}

\frac{55 x ^{7} y ^{6}}{11 x ^{11} y ^{8}}

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