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

The solutions to the equation 17^2=x^2+(x+7)^2 are x=8,x=-15

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

The zeros of 17^2=x^2+(x+7)^2 are x=8,x=-15

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1 7^{2} = x^{2} + (x + 7)^{2}

x = 8, x = - 15

Solution steps

1 7^{2} = x^{2} + (x + 7)^{2}

Expand

1 7^{2} : 289

Expand

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

289 = 2 x^{2} + 14 x + 49

Switch sides

2 x^{2} + 14 x + 49 = 289

Move

289

to the left side

2 x^{2} + 14 x - 240 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 ( - 240 )}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 (- 240) = 46

x_{1, 2} = \frac{- 14 \pm 46}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 14 + 46}{2 \cdot 2}, x_{2} = \frac{- 14 - 46}{2 \cdot 2}

x = \frac{- 14 + 46}{2 \cdot 2} : 8

x = \frac{- 14 - 46}{2 \cdot 2} : - 15

The solutions to the quadratic equation are:

x = 8, x = - 15

t, t^{2} + 12 t + a = 0

x^{2} - 11 x + 25 = 0

4 h^{2} = 3 l^{2}

x, z = x^{2} y - x + y + 1

D (x) = x^{2} - 6 x, - 2 \leq x < 3

What are the solutions to the equation 17^2=x^2+(x+7)^2 ?
The solutions to the equation 17^2=x^2+(x+7)^2 are x=8,x=-15
Find the zeros of 17^2=x^2+(x+7)^2
The zeros of 17^2=x^2+(x+7)^2 are x=8,x=-15