Maine MTA High School (grade 10) Algebra. Practice it free.

The equation means $x - 6 = 18$ or $x - 6 = -18$. So $x = 24$ or $x = -12$. Both solutions are $-12$ and $24\textsf{.}$

Either $x - 7 = 15$, giving $x = 22$, or $x - 7 = -15$, giving $x = -8$. The positive value of $x$ is $22\textsf{.}$

$2! = 2 \times 1 = 2$.

The greatest common factor of $14x^{2}$ and $21x$ is $7x\textsf{.}$ Factoring gives $7x(2x + 3)$.

At least $25$ and no more than $55$ include both endpoints, so $25 \le s \le 55$.

$x^{2} - 64 = x^{2} - 8^{2} = (x - 8)(x + 8)$.

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

Here $a = 1$, $b = 1$, and $c = 1$. Substituting into $b^{2} - 4ac$ gives $1^{2} - 4(1)(1) = -3$, so there are $0$ real solutions.