What are the solutions to the equation (x-2)(2x+4)=12 ?

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

Find the zeros of (x-2)(2x+4)=12

The zeros of (x-2)(2x+4)=12 are x=sqrt(10),x=-sqrt(10)

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(x - 2) (2 x + 4) = 12

x = 10, x = - 10

Decimal Notation

x = 3.16227 \dots, x = - 3.16227 \dots

Solution steps

(x - 2) (2 x + 4) = 12

Expand

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

2 x^{2} - 8 = 12

Move

12

to the left side

2 x^{2} - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

0^{2} - 4 \cdot 2 (- 20) = 410

x_{1, 2} = \frac{- 0 \pm 4 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 0 + 4 10}{2 \cdot 2}, x_{2} = \frac{- 0 - 4 10}{2 \cdot 2}

x = \frac{- 0 + 4 10}{2 \cdot 2} : 10

x = \frac{- 0 - 4 10}{2 \cdot 2} : - 10

The solutions to the quadratic equation are:

x = 10, x = - 10

2 x^{2} - 1 = 11 x + 5

3 x^{2} - 12 x + 13

(100 - 2 x) (50 - 2 x) = 3600

(c + 1)^{2} = \frac{21}{2}

x^{2} + 8 = - 6 x

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