What are the solutions to the equation x^2+(x+7)^2=(x+8)^2 ?

The solutions to the equation x^2+(x+7)^2=(x+8)^2 are x=5,x=-3

Find the zeros of x^2+(x+7)^2=(x+8)^2

The zeros of x^2+(x+7)^2=(x+8)^2 are x=5,x=-3

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x^{2} + (x + 7)^{2} = (x + 8)^{2}

x = 5, x = - 3

Solution steps

x^{2} + (x + 7)^{2} = (x + 8)^{2}

Expand

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

Expand

(x + 8)^{2} : x^{2} + 16 x + 64

2 x^{2} + 14 x + 49 = x^{2} + 16 x + 64

Move

64

to the left side

2 x^{2} + 14 x - 15 = x^{2} + 16 x

Move

16 x

to the left side

2 x^{2} - 2 x - 15 = x^{2}

Move

x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 1}

x = \frac{- ( - 2 ) + 8}{2 \cdot 1} : 5

x = \frac{- ( - 2 ) - 8}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 5, x = - 3

\frac{- 9 r ^{2} s ^{6} t ^{4}}{54 r ^{5} s ^{2} t ^{8}}

5 = 10 n

\frac{( z - 1 )}{3} - \frac{( z + 3 )}{2} = 5 z

5 y = 13

(x + 1) (x - 7) \leq 0

What are the solutions to the equation x^2+(x+7)^2=(x+8)^2 ?
The solutions to the equation x^2+(x+7)^2=(x+8)^2 are x=5,x=-3
Find the zeros of x^2+(x+7)^2=(x+8)^2
The zeros of x^2+(x+7)^2=(x+8)^2 are x=5,x=-3