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

The solutions to the equation y^2-2y-8=0 are y=4,y=-2

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

The zeros of y^2-2y-8=0 are y=4,y=-2

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

y = 4, y = - 2

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

y_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 1}, y_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 1}

y = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

y = \frac{- ( - 2 ) - 6}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

y = 4, y = - 2

- 6 \leq - 2 x + 2 < 8

5 (x^{2} - 10) + 5 = - 10

c + 5 = 5

x^{2} + 25 x = 0

\frac{x + 4}{6} + \frac{3 - 5 x}{5} = \frac{x - 1}{4}

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