What are the solutions to the equation 2w^2+18w+29=(w+6)^2 ?

The solutions to the equation 2w^2+18w+29=(w+6)^2 are w=1,w=-7

Find the zeros of 2w^2+18w+29=(w+6)^2

The zeros of 2w^2+18w+29=(w+6)^2 are w=1,w=-7

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2 w^{2} + 18 w + 29 = (w + 6)^{2}

w = 1, w = - 7

Solution steps

2 w^{2} + 18 w + 29 = (w + 6)^{2}

Expand

(w + 6)^{2} : w^{2} + 12 w + 36

2 w^{2} + 18 w + 29 = w^{2} + 12 w + 36

Move

36

to the left side

2 w^{2} + 18 w - 7 = w^{2} + 12 w

Move

12 w

to the left side

2 w^{2} + 6 w - 7 = w^{2}

Move

w^{2}

to the left side

w^{2} + 6 w - 7 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 7) = 8

w_{1, 2} = \frac{- 6 \pm 8}{2 \cdot 1}

Separate the solutions

w_{1} = \frac{- 6 + 8}{2 \cdot 1}, w_{2} = \frac{- 6 - 8}{2 \cdot 1}

w = \frac{- 6 + 8}{2 \cdot 1} : 1

w = \frac{- 6 - 8}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

w = 1, w = - 7

121 x^{2} - 144

\frac{( 4 ^{- 0.1} - 1 )}{- 0.1}

- 24 = 48 - 2 (x + 4)^{2}

1 2^{2} + b^{2} = 1 5^{2}

∣ 4 x - 7 ∣ + 3 = 7

What are the solutions to the equation 2w^2+18w+29=(w+6)^2 ?
The solutions to the equation 2w^2+18w+29=(w+6)^2 are w=1,w=-7
Find the zeros of 2w^2+18w+29=(w+6)^2
The zeros of 2w^2+18w+29=(w+6)^2 are w=1,w=-7