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

The solutions to the equation y^2-4y+2=0 are y=2+sqrt(2),y=2-sqrt(2)

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

The zeros of y^2-4y+2=0 are y=2+sqrt(2),y=2-sqrt(2)

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

y = 2 + 2, y = 2 - 2

Decimal Notation

y = 3.41421 \dots, y = 0.58578 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 2 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 + 2, y = 2 - 2

x + 10 = 2 x + 4

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

3 x^{2} - 7 x - 9 = 4 x^{2} - 4 x - 9

a^{2} - 1

- 3 (6 + 3 x) = - 72

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