MCA High School / MCA EOC Algebra. Practice it free.

The equation means $3x - 12 = 24$ or $3x - 12 = -24$. Adding $12$ gives $3x = 36$ or $3x = -12$. Dividing by $3$ gives $x = 12$ or $x = -4$. Both solutions are $-4$ and $12\textsf{.}$

Rewrite $|6x + 12| = |6(x + 2)| = |6|\,|x + 2| = 6|x + 2|$. Since $6|x + 2| = 48$, divide by $6$ to get $|x + 2| = 8$. The expression $x + 2$ equals $8$ or $-8$; the positive value is $8\textsf{.}$

The greatest common factor of $20x^{2}$ and $30x$ is $10x\textsf{.}$ Factoring gives $10x(2x - 3)$.

At least $2$ gives $d \ge 2$. Less than $7$ gives $d < 7$, so $2 \le d < 7$.

$4x^{2} - 9 = (2x)^{2} - 3^{2} = (2x - 3)(2x + 3)$.

Here $a = 1$, $b = 0$, and $c = -49$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-49) = 196\textsf{,}$ so there are $2$ distinct real roots.

Here $a = 1$, $b = 0$, and $c = -16$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-16) = 64$.

Write $30u$ with denominator $g$: $\dfrac{10u}{g} + \dfrac{30ug}{g} = \dfrac{10u + 30ug}{g} = \dfrac{10u(1 + 3g)}{g}$.