What are the solutions to the equation x^2+(x+6)^2=(x+7)^2 ?

The solutions to the equation x^2+(x+6)^2=(x+7)^2 are x=1+sqrt(14),x=1-sqrt(14)

Find the zeros of x^2+(x+6)^2=(x+7)^2

The zeros of x^2+(x+6)^2=(x+7)^2 are x=1+sqrt(14),x=1-sqrt(14)

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x^{2} + (x + 6)^{2} = (x + 7)^{2}

x = 1 + 14, x = 1 - 14

Solution steps

x^{2} + (x + 6)^{2} = (x + 7)^{2}

Expand

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

Expand

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

2 x^{2} + 12 x + 36 = x^{2} + 14 x + 49

Move

49

to the left side

2 x^{2} + 12 x - 13 = x^{2} + 14 x

Move

14 x

to the left side

2 x^{2} - 2 x - 13 = x^{2}

Move

x^{2}

to the left side

x^{2} - 2 x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 13) = 214

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 2 14}{2 \cdot 1} : 1 + 14

x = \frac{- ( - 2 ) - 2 14}{2 \cdot 1} : 1 - 14

The solutions to the quadratic equation are:

x = 1 + 14, x = 1 - 14

- 4 (y + 1) + 6 y = 2 (y + 2) - 8

9 x^{2} + 6 x + 1

7 - 3 x \leq - 20 or 5 x - 6 \leq 9

\frac{b ^{- 4}}{b ^{- 5}}

0 = x^{2} + 2 x + 4

What are the solutions to the equation x^2+(x+6)^2=(x+7)^2 ?
The solutions to the equation x^2+(x+6)^2=(x+7)^2 are x=1+sqrt(14),x=1-sqrt(14)
Find the zeros of x^2+(x+6)^2=(x+7)^2
The zeros of x^2+(x+6)^2=(x+7)^2 are x=1+sqrt(14),x=1-sqrt(14)