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

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

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

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

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

y = 2, y = - 4

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} : 2

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

The solutions to the quadratic equation are:

y = 2, y = - 4

6 x + 35 = - 6 x + 35

14 \cdot \frac{2}{3}

\frac{1}{8} + \frac{1}{4} + \frac{1}{2}

(\frac{6}{n})^{- 2}

∣ 6 x ∣ = 18

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