What are the solutions to the equation p^2+(p+7)^2=169 ?

The solutions to the equation p^2+(p+7)^2=169 are p=5,p=-12

Find the zeros of p^2+(p+7)^2=169

The zeros of p^2+(p+7)^2=169 are p=5,p=-12

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p^{2} + (p + 7)^{2} = 169

p = 5, p = - 12

Solution steps

p^{2} + (p + 7)^{2} = 169

Expand

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

2 p^{2} + 14 p + 49 = 169

Move

169

to the left side

2 p^{2} + 14 p - 120 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 ( - 120 )}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 (- 120) = 34

p_{1, 2} = \frac{- 14 \pm 34}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- 14 + 34}{2 \cdot 2}, p_{2} = \frac{- 14 - 34}{2 \cdot 2}

p = \frac{- 14 + 34}{2 \cdot 2} : 5

p = \frac{- 14 - 34}{2 \cdot 2} : - 12

The solutions to the quadratic equation are:

p = 5, p = - 12

4 n^{2} + 23 n + 16 = 1

- 1 L^{2} + 41.5 L = 0

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

x^{2} + 18 = - 12 x

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

What are the solutions to the equation p^2+(p+7)^2=169 ?
The solutions to the equation p^2+(p+7)^2=169 are p=5,p=-12
Find the zeros of p^2+(p+7)^2=169
The zeros of p^2+(p+7)^2=169 are p=5,p=-12