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

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

Find the zeros of y^2+2y=48

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

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y^{2} + 2 y = 48

y = 6, y = - 8

Solution steps

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

5 ∣ 2 x - 1 ∣ + 3 = - ∣ x - 3 ∣ + 10

5 x = - 2 x + 28

1 - 100 x^{2}

2 d^{2} - d - 15

(7 x - 3)^{2} = 9

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