What are the solutions to the equation (y+5)(y-1)=27 ?

The solutions to the equation (y+5)(y-1)=27 are y=4,y=-8

Find the zeros of (y+5)(y-1)=27

The zeros of (y+5)(y-1)=27 are y=4,y=-8

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(y + 5) (y - 1) = 27

y = 4, y = - 8

Solution steps

(y + 5) (y - 1) = 27

Expand

(y + 5) (y - 1) : y^{2} + 4 y - 5

y^{2} + 4 y - 5 = 27

Move

27

to the left side

y^{2} + 4 y - 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

y_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 12}{2 \cdot 1}, y_{2} = \frac{- 4 - 12}{2 \cdot 1}

y = \frac{- 4 + 12}{2 \cdot 1} : 4

y = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = 4, y = - 8

x^{2} - 5 x + 20 = 0

- 4 x^{2} + 56 x + 0 = 0

\frac{- 1 - 2 i}{- 1 + 2 i}

(1 + \frac{0.06}{4})^{4}

\frac{1}{2 + 3}

What are the solutions to the equation (y+5)(y-1)=27 ?
The solutions to the equation (y+5)(y-1)=27 are y=4,y=-8
Find the zeros of (y+5)(y-1)=27
The zeros of (y+5)(y-1)=27 are y=4,y=-8