What are the solutions to the equation 2y^2-8y+7=0 ?

The solutions to the equation 2y^2-8y+7=0 are y=(4+sqrt(2))/2 ,y=(4-sqrt(2))/2

Find the zeros of 2y^2-8y+7=0

The zeros of 2y^2-8y+7=0 are y=(4+sqrt(2))/2 ,y=(4-sqrt(2))/2

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2 y^{2} - 8 y + 7 = 0

y = \frac{4 + 2}{2}, y = \frac{4 - 2}{2}

Decimal Notation

y = 2.70710 \dots, y = 1.29289 \dots

Solution steps

2 y^{2} - 8 y + 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 7 = 22

y_{1, 2} = \frac{- ( - 8 ) \pm 2 2}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 2 2}{2 \cdot 2}, y_{2} = \frac{- ( - 8 ) - 2 2}{2 \cdot 2}

y = \frac{- ( - 8 ) + 2 2}{2 \cdot 2} : \frac{4 + 2}{2}

y = \frac{- ( - 8 ) - 2 2}{2 \cdot 2} : \frac{4 - 2}{2}

The solutions to the quadratic equation are:

y = \frac{4 + 2}{2}, y = \frac{4 - 2}{2}

x, x^{2} + y^{2} = - 4

x^{2} + 6 x + 10 = 0

30 x^{2} - 26 x + 4 = 0

6 x^{2} + 7 = 0

x, 48 = x^{2} + 4 x y

What are the solutions to the equation 2y^2-8y+7=0 ?
The solutions to the equation 2y^2-8y+7=0 are y=(4+sqrt(2))/2 ,y=(4-sqrt(2))/2
Find the zeros of 2y^2-8y+7=0
The zeros of 2y^2-8y+7=0 are y=(4+sqrt(2))/2 ,y=(4-sqrt(2))/2