What are the solutions to the equation y^2-8y-4=0 ?

The solutions to the equation y^2-8y-4=0 are y=2(2+sqrt(5)),y=2(2-sqrt(5))

Find the zeros of y^2-8y-4=0

The zeros of y^2-8y-4=0 are y=2(2+sqrt(5)),y=2(2-sqrt(5))

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y^{2} - 8 y - 4 = 0

y = 2 (2 + 5), y = 2 (2 - 5)

Decimal Notation

y = 8.47213 \dots, y = - 0.47213 \dots

Solution steps

y^{2} - 8 y - 4 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 1 \cdot (- 4) = 45

y_{1, 2} = \frac{- ( - 8 ) \pm 4 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 4 5}{2 \cdot 1}, y_{2} = \frac{- ( - 8 ) - 4 5}{2 \cdot 1}

y = \frac{- ( - 8 ) + 4 5}{2 \cdot 1} : 2 (2 + 5)

y = \frac{- ( - 8 ) - 4 5}{2 \cdot 1} : 2 (2 - 5)

The solutions to the quadratic equation are:

y = 2 (2 + 5), y = 2 (2 - 5)

b^{2} - 10 b + 21 = 0

x (2 x - 3) = 18

- 5 x^{2} + 60 x - 405 = 0

- 3 b^{2} + b - A = 0

4 x^{2} - 6 x + 5 = x + 1

What are the solutions to the equation y^2-8y-4=0 ?
The solutions to the equation y^2-8y-4=0 are y=2(2+sqrt(5)),y=2(2-sqrt(5))
Find the zeros of y^2-8y-4=0
The zeros of y^2-8y-4=0 are y=2(2+sqrt(5)),y=2(2-sqrt(5))