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

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

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

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

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2 y + y^{2} = 15

y = 3, y = - 5

Solution steps

2 y + y^{2} = 15

Move

15

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

- 2 x + 7 < 23

7 b + 6 b^{2}

3 - 412

m - 88 < - 176

(- \frac{3}{7})^{2} + y^{2} = 1

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