What are the solutions to the equation 2y^2+17y+45=(y+7)^2 ?

The solutions to the equation 2y^2+17y+45=(y+7)^2 are y=1,y=-4

Find the zeros of 2y^2+17y+45=(y+7)^2

The zeros of 2y^2+17y+45=(y+7)^2 are y=1,y=-4

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2 y^{2} + 17 y + 45 = (y + 7)^{2}

y = 1, y = - 4

Solution steps

2 y^{2} + 17 y + 45 = (y + 7)^{2}

Expand

(y + 7)^{2} : y^{2} + 14 y + 49

2 y^{2} + 17 y + 45 = y^{2} + 14 y + 49

Move

49

to the left side

2 y^{2} + 17 y - 4 = y^{2} + 14 y

Move

14 y

to the left side

2 y^{2} + 3 y - 4 = y^{2}

Move

y^{2}

to the left side

y^{2} + 3 y - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

y_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 5}{2 \cdot 1}, y_{2} = \frac{- 3 - 5}{2 \cdot 1}

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

y = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

y = 1, y = - 4

∣ x ∣ = 20

p^{3} + 27

- 3 < 3

\frac{1530}{115000} \cdot \frac{2}{2}

\frac{x - 2}{- 2} = 6

What are the solutions to the equation 2y^2+17y+45=(y+7)^2 ?
The solutions to the equation 2y^2+17y+45=(y+7)^2 are y=1,y=-4
Find the zeros of 2y^2+17y+45=(y+7)^2
The zeros of 2y^2+17y+45=(y+7)^2 are y=1,y=-4