What are the solutions to the equation y(y+2)=48 ?

The solutions to the equation y(y+2)=48 are y=6,y=-8

Find the zeros of y(y+2)=48

The zeros of y(y+2)=48 are y=6,y=-8

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y (y + 2) = 48

y = 6, y = - 8

Solution steps

y (y + 2) = 48

Expand

y (y + 2) : y^{2} + 2 y

y^{2} + 2 y = 48

Move

48

to the left side

y^{2} + 2 y - 48 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 48) = 14

y_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 2 + 14}{2 \cdot 1}, y_{2} = \frac{- 2 - 14}{2 \cdot 1}

y = \frac{- 2 + 14}{2 \cdot 1} : 6

y = \frac{- 2 - 14}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = 6, y = - 8

3 y^{2} - 2 y - 1 = 0

4 x^{2} + 4 x^{2} = 36

x^{2} - 6 x = 16

(5 p + 3)^{2} = 12

x^{2} + 6 x = 5, x < - 1

What are the solutions to the equation y(y+2)=48 ?
The solutions to the equation y(y+2)=48 are y=6,y=-8
Find the zeros of y(y+2)=48
The zeros of y(y+2)=48 are y=6,y=-8