What are the solutions to the equation (5-y+1)^2+(y+1)^2=25 ?

The solutions to the equation (5-y+1)^2+(y+1)^2=25 are y=3,y=2

Find the zeros of (5-y+1)^2+(y+1)^2=25

The zeros of (5-y+1)^2+(y+1)^2=25 are y=3,y=2

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(5 - y + 1)^{2} + (y + 1)^{2} = 25

y = 3, y = 2

Solution steps

(5 - y + 1)^{2} + (y + 1)^{2} = 25

Expand

(5 - y + 1)^{2} + (y + 1)^{2} : 2 y^{2} - 10 y + 37

2 y^{2} - 10 y + 37 = 25

Move

25

to the left side

2 y^{2} - 10 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 12}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 \cdot 12 = 2

y_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 2}, y_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 2}

y = \frac{- ( - 10 ) + 2}{2 \cdot 2} : 3

y = \frac{- ( - 10 ) - 2}{2 \cdot 2} : 2

The solutions to the quadratic equation are:

y = 3, y = 2

9 y^{2} + 24 y + 16 = 0

2 x^{2} + 12 x = 18

- 15 + 18 x - 3 x^{2} = 0

x^{2} + 3 a x = 6 a^{2}

7 s^{2} + 9 s - 4 = 0

What are the solutions to the equation (5-y+1)^2+(y+1)^2=25 ?
The solutions to the equation (5-y+1)^2+(y+1)^2=25 are y=3,y=2
Find the zeros of (5-y+1)^2+(y+1)^2=25
The zeros of (5-y+1)^2+(y+1)^2=25 are y=3,y=2