Find a,b,c (x-2)(2x+4)=12

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

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $(x - 2) (2 x + 4)$ .

Expand $(x - 2) (2 x + 4)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$x (2 x) + x \cdot 4 - 2 (2 x + 4) = 12$

Apply the distributive property.

$x (2 x) + x \cdot 4 - 2 (2 x) - 2 \cdot 4 = 12$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot 4 - 2 (2 x) - 2 \cdot 4 = 12$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) + x \cdot 4 - 2 (2 x) - 2 \cdot 4 = 12$

Multiply $x$ by $x$ .

$2 x^{2} + x \cdot 4 - 2 (2 x) - 2 \cdot 4 = 12$

Move $4$ to the left of $x$ .

$2 x^{2} + 4 \cdot x - 2 (2 x) - 2 \cdot 4 = 12$

Multiply $2$ by $- 2$ .

$2 x^{2} + 4 x - 4 x - 2 \cdot 4 = 12$

Multiply $- 2$ by $4$ .

$2 x^{2} + 4 x - 4 x - 8 = 12$

Subtract $4 x$ from $4 x$ .

$2 x^{2} + 0 - 8 = 12$

Add $2 x^{2}$ and $0$ .

$2 x^{2} - 8 = 12$

Subtract $12$ from both sides of the equation.

$2 x^{2} - 8 - 12 = 0$

Subtract $12$ from $- 8$ .

$2 x^{2} - 20 = 0$

Once the quadratic is in standard form, the values of $a$ , $b$ , and $c$ can be found.

$a x^{2} + b x + c$

Use the standard form of the equation to find $a$ , $b$ , and $c$ for this quadratic.

$a = 2$ , $b = 0$ , $c = - 20$