What are the solutions to the equation (y-2)(y+5)=(2y-1)(y+1)+7 ?

The solutions to the equation (y-2)(y+5)=(2y-1)(y+1)+7 are y=1-sqrt(15)i,y=1+sqrt(15)i

Find the zeros of (y-2)(y+5)=(2y-1)(y+1)+7

The zeros of (y-2)(y+5)=(2y-1)(y+1)+7 are y=1-sqrt(15)i,y=1+sqrt(15)i

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(y - 2) (y + 5) = (2 y - 1) (y + 1) + 7

y = 1 - 15 i, y = 1 + 15 i

Solution steps

(y - 2) (y + 5) = (2 y - 1) (y + 1) + 7

Expand

(y - 2) (y + 5) : y^{2} + 3 y - 10

Expand

(2 y - 1) (y + 1) + 7 : 2 y^{2} + y + 6

y^{2} + 3 y - 10 = 2 y^{2} + y + 6

Move

6

to the left side

y^{2} + 3 y - 16 = 2 y^{2} + y

Move

y

to the left side

y^{2} + 2 y - 16 = 2 y^{2}

Move

2 y^{2}

to the left side

- y^{2} + 2 y - 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) ( - 16 )}{2 ( - 1 )}

Simplify

2^{2} - 4 (- 1) (- 16) : 215 i

y_{1, 2} = \frac{- 2 \pm 2 15 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 2 + 2 15 i}{2 ( - 1 )}, y_{2} = \frac{- 2 - 2 15 i}{2 ( - 1 )}

y = \frac{- 2 + 2 15 i}{2 ( - 1 )} : 1 - 15 i

y = \frac{- 2 - 2 15 i}{2 ( - 1 )} : 1 + 15 i

The solutions to the quadratic equation are:

y = 1 - 15 i, y = 1 + 15 i

x^{2} + 10 x + 24 = 0

x^{2} + 13 x

t^{2} - 8 t + 16 = 0

x^{2} + 4 x = - 9

x^{2} - 19 x + 90 = 0

What are the solutions to the equation (y-2)(y+5)=(2y-1)(y+1)+7 ?
The solutions to the equation (y-2)(y+5)=(2y-1)(y+1)+7 are y=1-sqrt(15)i,y=1+sqrt(15)i
Find the zeros of (y-2)(y+5)=(2y-1)(y+1)+7
The zeros of (y-2)(y+5)=(2y-1)(y+1)+7 are y=1-sqrt(15)i,y=1+sqrt(15)i