What are the solutions to the equation y^2+2y-15=0 ?

The solutions to the equation y^2+2y-15=0 are y=3,y=-5

Find the zeros of y^2+2y-15=0

The zeros of y^2+2y-15=0 are y=3,y=-5

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y^{2} + 2 y - 15 = 0

y = 3, y = - 5

Solution steps

y^{2} + 2 y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

y_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 2 + 8}{2 \cdot 1}, y_{2} = \frac{- 2 - 8}{2 \cdot 1}

y = \frac{- 2 + 8}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

y = 3, y = - 5

\frac{x ^{- 2}}{x ^{3}}

39

- 10 (3 + x) < - 110

\frac{- 9 - ( - 3 )}{8 - ( - 4 )}

- 4 + (- 2)

What are the solutions to the equation y^2+2y-15=0 ?
The solutions to the equation y^2+2y-15=0 are y=3,y=-5
Find the zeros of y^2+2y-15=0
The zeros of y^2+2y-15=0 are y=3,y=-5