What are the solutions to the equation (3-y)(y+4)=3y-5 ?

The solutions to the equation (3-y)(y+4)=3y-5 are y=-2-sqrt(21),y=sqrt(21)-2

Find the zeros of (3-y)(y+4)=3y-5

The zeros of (3-y)(y+4)=3y-5 are y=-2-sqrt(21),y=sqrt(21)-2

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(3 - y) (y + 4) = 3 y - 5

y = - 2 - 21, y = 21 - 2

Solution steps

(3 - y) (y + 4) = 3 y - 5

Expand

(3 - y) (y + 4) : - y - y^{2} + 12

- y - y^{2} + 12 = 3 y - 5

Move

5

to the left side

- y^{2} - y + 17 = 3 y

Move

3 y

to the left side

- y^{2} - 4 y + 17 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) \cdot 17}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) \cdot 17 = 221

y_{1, 2} = \frac{- ( - 4 ) \pm 2 21}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 2 21}{2 ( - 1 )}, y_{2} = \frac{- ( - 4 ) - 2 21}{2 ( - 1 )}

y = \frac{- ( - 4 ) + 2 21}{2 ( - 1 )} : - 2 - 21

y = \frac{- ( - 4 ) - 2 21}{2 ( - 1 )} : 21 - 2

The solutions to the quadratic equation are:

y = - 2 - 21, y = 21 - 2

3 + 4

220 + 845 - 80

2 ∣ 25 x ∣ = 100

6 + 7

\frac{8}{18088} \cdot 100

What are the solutions to the equation (3-y)(y+4)=3y-5 ?
The solutions to the equation (3-y)(y+4)=3y-5 are y=-2-sqrt(21),y=sqrt(21)-2
Find the zeros of (3-y)(y+4)=3y-5
The zeros of (3-y)(y+4)=3y-5 are y=-2-sqrt(21),y=sqrt(21)-2