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=16,x=-12

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 16, x = - 12

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} : 16

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

The solutions to the quadratic equation are:

x = 16, x = - 12

44 = - 16 t^{2} + 64 t + 80

3 c^{2} - 2 c - 2 = 0

0 = 8 x^{2}

(x + 1)^{2} - 3 = 13

3 x^{2} - 20 x + 31 = 0

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