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

The solutions to the equation y^2-4y-8=0 are y=2(1+sqrt(3)),y=2(1-sqrt(3))

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

The zeros of y^2-4y-8=0 are y=2(1+sqrt(3)),y=2(1-sqrt(3))

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

y = 2 (1 + 3), y = 2 (1 - 3)

Decimal Notation

y = 5.46410 \dots, y = - 1.46410 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 8) = 43

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 (1 + 3), y = 2 (1 - 3)

b, b^{2} - m^{2} = 0

4 x^{2} + 53 = 28 x

9 z^{2} - 27 z + 20 = 0

y, x^{2} + y^{2} - x y = 1

y, x^{2} + 2 x y - y^{2} + x = 36

What are the solutions to the equation y^2-4y-8=0 ?
The solutions to the equation y^2-4y-8=0 are y=2(1+sqrt(3)),y=2(1-sqrt(3))
Find the zeros of y^2-4y-8=0
The zeros of y^2-4y-8=0 are y=2(1+sqrt(3)),y=2(1-sqrt(3))