What are the solutions to the equation (2x-4)^2-2x(x-2)=48 ?

The solutions to the equation (2x-4)^2-2x(x-2)=48 are x=8,x=-2

Find the zeros of (2x-4)^2-2x(x-2)=48

The zeros of (2x-4)^2-2x(x-2)=48 are x=8,x=-2

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(2 x - 4)^{2} - 2 x (x - 2) = 48

x = 8, x = - 2

Solution steps

(2 x - 4)^{2} - 2 x (x - 2) = 48

Expand

(2 x - 4)^{2} - 2 x (x - 2) : 2 x^{2} - 12 x + 16

2 x^{2} - 12 x + 16 = 48

Move

48

to the left side

2 x^{2} - 12 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 ( - 32 )}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 (- 32) = 20

x_{1, 2} = \frac{- ( - 12 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 20}{2 \cdot 2}

x = \frac{- ( - 12 ) + 20}{2 \cdot 2} : 8

x = \frac{- ( - 12 ) - 20}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = 8, x = - 2

4 \cdot (3 - x) < x + 12

\frac{3}{4} + 1

4 - 64

3 ∣ x - 2 ∣ + 6 > 9

x^{2} - 2 x > 9

What are the solutions to the equation (2x-4)^2-2x(x-2)=48 ?
The solutions to the equation (2x-4)^2-2x(x-2)=48 are x=8,x=-2
Find the zeros of (2x-4)^2-2x(x-2)=48
The zeros of (2x-4)^2-2x(x-2)=48 are x=8,x=-2