Let's expand: ${(3x-2)}^2$

ⓘ Raising a sum to an integer power is multiplying it by itself: ${(3x-2)}^2$ = $(3x-2)·(3x-2)$

$(3x-2)·(3x-2)$

ⓘ Expand by multiplying each term inside the parentheses: $(3x-2)·(3x-2)$ = $3·x·3·x$$+3x·\left(-2\right)$$-2·3x$$-2·\left(-2\right)$

$3·x·3·x$$+3x·\left(-2\right)$$-2·3x$$-2·\left(-2\right)$

ⓘ Multiplications are done before additions and subtractions: $3·3$ = $9$, $2·3$ = $6$■Multiplying powers with the same base adds the exponents: $x·x$ = $x^2$■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative: $3·\left(-2\right)$ = $-6$■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: $-2·\left(-2\right)$ = $4$

$9x^2$$-6x$$-6x$$+4$

ⓘ Like terms are grouped by adding their coefficients: $-6x-6x$ = $-12x$

▶Expanded form:   $9x^2-12x+4$