Simplify the following expressions: a) (z-7) b1z-a c) z-12

(a) $(z-1)^2 - 3(z-3)$

Step 1: Expand $(z-1)^2$.
$(z-1)^2 = z^2 - 2z + 1$

Step 2: Expand $3(z-3)$.
$3(z-3) = 3z - 9$

Step 3: Subtract.
$z^2 - 2z + 1 - (3z - 9) = z^2 - 2z + 1 - 3z + 9$
$= z^2 - 5z + 10$

Simplified: $z^2 - 5z + 10$
$\boxed{z^2 - 5z + 10}$

(b) $x^2 - 5x + 6 = 0$

Step 1: Sum of roots formula: $-\frac{b}{a}$.
$a=1$, $b=-5$
$-\frac{-5}{1} = 5$

Step 2: Product of roots formula: $\frac{c}{a}$.
$c=6$
$\frac{6}{1} = 6$

Sum $= 5$, Product $= 6$
$\boxed{Sum = 5, Product = 6}$

(c) $2x^2 + 3x - 2 = 0$

Step 1: Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
$a=2$, $b=3$, $c=-2$

Step 2: Discriminant $D = b^2 - 4ac$.
$D = 3^2 - 4(2)(-2) = 9 + 16 = 25$
$\sqrt{D} = 5$

Step 3: Roots.
$x = \frac{-3 \pm 5}{4}$
$x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$
$x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$

Roots: $x = \frac{1}{2}, -2$
$\boxed{x = \dfrac{1}{2}, -2}$

(d) $x^2 + 4x + 4 = 0$

Step 1: Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
$a=1$, $b=4$, $c=4$

Step 2: Discriminant $D = b^2 - 4ac$.
$D = 16 - 16 = 0$
$\sqrt{D} = 0$

Step 3: Root.
$x = \frac{-4 \pm 0}{2} = \frac{-4}{2} = -2$

Root: $x = -2$ (repeated)
$\boxed{x = -2}$

(e) $3x^2 + 2x - 1 = 0$

Step 1: Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
$a=3$, $b=2$, $c=-1$

Step 2: Discriminant $D = b^2 - 4ac$.
$D = 4 - 4(3)(-1) = 4 + 12 = 16$
$\sqrt{D} = 4$

Step 3: Roots.
$x = \frac{-2 \pm 4}{6}$
$x_1 = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$
$x_2 = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$

Roots: $x = \frac{1}{3}, -1$
$\boxed{x = \dfrac{1}{3}, -1}$