What are the solutions to the equation y^2+3y=28 ?

The solutions to the equation y^2+3y=28 are y=4,y=-7

Find the zeros of y^2+3y=28

The zeros of y^2+3y=28 are y=4,y=-7

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y^{2} + 3 y = 28

y = 4, y = - 7

Solution steps

y^{2} + 3 y = 28

Move

28

to the left side

y^{2} + 3 y - 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

y_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 11}{2 \cdot 1}, y_{2} = \frac{- 3 - 11}{2 \cdot 1}

y = \frac{- 3 + 11}{2 \cdot 1} : 4

y = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

y = 4, y = - 7

2 k (k + 3) = - 9

2 x^{2} + 8 x - 1 = 0

2 x^{2} - 2 x - 10 = 0

4 x^{2} + 5 = 12

\frac{1}{8} x^{2} + \frac{5}{4} x + \frac{1}{3} = 0

What are the solutions to the equation y^2+3y=28 ?
The solutions to the equation y^2+3y=28 are y=4,y=-7
Find the zeros of y^2+3y=28
The zeros of y^2+3y=28 are y=4,y=-7