What are the solutions to the equation 29-Q^2=(Q+3)^2 ?

The solutions to the equation 29-Q^2=(Q+3)^2 are Q=2,Q=-5

Find the zeros of 29-Q^2=(Q+3)^2

The zeros of 29-Q^2=(Q+3)^2 are Q=2,Q=-5

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29 - Q^{2} = (Q + 3)^{2}

Q = 2, Q = - 5

Solution steps

29 - Q^{2} = (Q + 3)^{2}

Expand

(Q + 3)^{2} : Q^{2} + 6 Q + 9

29 - Q^{2} = Q^{2} + 6 Q + 9

Switch sides

Q^{2} + 6 Q + 9 = 29 - Q^{2}

Move

Q^{2}

to the left side

2 Q^{2} + 6 Q + 9 = 29

Move

29

to the left side

2 Q^{2} + 6 Q - 20 = 0

Solve with the quadratic formula

Q_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 20) = 14

Q_{1, 2} = \frac{- 6 \pm 14}{2 \cdot 2}

Separate the solutions

Q_{1} = \frac{- 6 + 14}{2 \cdot 2}, Q_{2} = \frac{- 6 - 14}{2 \cdot 2}

Q = \frac{- 6 + 14}{2 \cdot 2} : 2

Q = \frac{- 6 - 14}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

Q = 2, Q = - 5

(\frac{2 x ^{2} y}{- 2 x ^{2} y ^{3}})^{- 1}

9 m^{2} - 16 n^{2}

6 p^{4} + 7 p^{2} - 3

z^{2} + 5 = 0

(x - 3)^{2} (18 - 3 x) > 0

What are the solutions to the equation 29-Q^2=(Q+3)^2 ?
The solutions to the equation 29-Q^2=(Q+3)^2 are Q=2,Q=-5
Find the zeros of 29-Q^2=(Q+3)^2
The zeros of 29-Q^2=(Q+3)^2 are Q=2,Q=-5