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

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

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

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

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

x = 16, x = - 12

Solution steps

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

Expand

2 0^{2} : 400

Expand

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

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

Switch sides

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

2 x^{2} - 13 x - 7 = 9 x

7 m^{2} - 24 m = - 9

t^{2} + 3 t - 19 = - 5

3 z^{2} - 7 z + 3 = 6

(- 4 - x) (1 - x) + 6 = 0

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