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

The solutions to the equation y^2+2y-8y=0 are y=6,y=0

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

The zeros of y^2+2y-8y=0 are y=6,y=0

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

y = 6, y = 0

Solution steps

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

Refine

y^{2} - 6 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 0 = 6

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

Separate the solutions

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

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

y = \frac{- ( - 6 ) - 6}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

y = 6, y = 0

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

A (x) = (x + 1)^{2} - 4

6 a^{2} x^{2} - 11 a x = 14

x^{2} - 6 x + 13 = 0

5 x^{2} - 16 x = - 2 x + 3

What are the solutions to the equation y^2+2y-8y=0 ?
The solutions to the equation y^2+2y-8y=0 are y=6,y=0
Find the zeros of y^2+2y-8y=0
The zeros of y^2+2y-8y=0 are y=6,y=0