What are the solutions to the equation 2y^2+y-9=(y-1)^2 ?

The solutions to the equation 2y^2+y-9=(y-1)^2 are y=2,y=-5

Find the zeros of 2y^2+y-9=(y-1)^2

The zeros of 2y^2+y-9=(y-1)^2 are y=2,y=-5

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2 y^{2} + y - 9 = (y - 1)^{2}

y = 2, y = - 5

Solution steps

2 y^{2} + y - 9 = (y - 1)^{2}

Expand

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

2 y^{2} + y - 9 = y^{2} - 2 y + 1

Move

1

to the left side

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

Move

2 y

to the left side

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

Move

y^{2}

to the left side

y^{2} + 3 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

y_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 7}{2 \cdot 1}, y_{2} = \frac{- 3 - 7}{2 \cdot 1}

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

y = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

y = 2, y = - 5

4 x^{2} - 8 x = - 19

16 m^{2} - 96 m + 176

5 s^{2} - 26 s = 24

x^{3} y^{3} + 8

2 (5 x + 2) = 50

What are the solutions to the equation 2y^2+y-9=(y-1)^2 ?
The solutions to the equation 2y^2+y-9=(y-1)^2 are y=2,y=-5
Find the zeros of 2y^2+y-9=(y-1)^2
The zeros of 2y^2+y-9=(y-1)^2 are y=2,y=-5