What are the solutions to the equation x^2+(x+4)^2=20^2 ?

The solutions to the equation x^2+(x+4)^2=20^2 are x=12,x=-16

Find the zeros of x^2+(x+4)^2=20^2

The zeros of x^2+(x+4)^2=20^2 are x=12,x=-16

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

x = 12, x = - 16

Solution steps

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

Expand

x^{2} + (x + 4)^{2} : 2 x^{2} + 8 x + 16

Expand

2 0^{2} : 400

2 x^{2} + 8 x + 16 = 400

Move

400

to the left side

2 x^{2} + 8 x - 384 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 384 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 384) = 56

x_{1, 2} = \frac{- 8 \pm 56}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 56}{2 \cdot 2}, x_{2} = \frac{- 8 - 56}{2 \cdot 2}

x = \frac{- 8 + 56}{2 \cdot 2} : 12

x = \frac{- 8 - 56}{2 \cdot 2} : - 16

The solutions to the quadratic equation are:

x = 12, x = - 16

5 (x - 6) + 3 = 18 - (3 + 2 x)

3 + 4 c = 15

\frac{8 x ^{2} y ^{2} - 10 x y ^{3} - 5 y}{2 y ^{2}}

- 2 x - 4 = - 16

3^{2} + 2^{3}

What are the solutions to the equation x^2+(x+4)^2=20^2 ?
The solutions to the equation x^2+(x+4)^2=20^2 are x=12,x=-16
Find the zeros of x^2+(x+4)^2=20^2
The zeros of x^2+(x+4)^2=20^2 are x=12,x=-16