What are the solutions to the equation x^2-(x-4)2=128 ?

The solutions to the equation x^2-(x-4)2=128 are x=12,x=-10

Find the zeros of x^2-(x-4)2=128

The zeros of x^2-(x-4)2=128 are x=12,x=-10

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x^{2} - (x - 4) 2 = 128

x = 12, x = - 10

Solution steps

x^{2} - (x - 4) \cdot 2 = 128

Expand

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

x^{2} - 2 x + 8 = 128

Move

128

to the left side

x^{2} - 2 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 120 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 120) = 22

x_{1, 2} = \frac{- ( - 2 ) \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 22}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 22}{2 \cdot 1}

x = \frac{- ( - 2 ) + 22}{2 \cdot 1} : 12

x = \frac{- ( - 2 ) - 22}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

x = 12, x = - 10

a^{2} + 1 = 2

5 x^{2} + 80 x = - 270

5 x^{2} - 10 x - 8 = 0

y, 2 x^{2} y + \frac{( x ^{3} - y ^{2} )}{x} + 3 = 0

- x^{2} + 3 x + 1 = 0

What are the solutions to the equation x^2-(x-4)2=128 ?
The solutions to the equation x^2-(x-4)2=128 are x=12,x=-10
Find the zeros of x^2-(x-4)2=128
The zeros of x^2-(x-4)2=128 are x=12,x=-10