\[(x-2)(x-5)=0\] \[x=2,x=5\]

\[(x-3)(x-4)=0\] \[x=3,x=4\]

\[(x+2)(x-10)=0\] \[x=-2,x=10\]

We look for two numbers that multiply to \(12\) and add up to \(-7\): \[ -3 \times -4 = 12 \quad \text{and} \quad -3 + -4 = -7 \] Factor: \[ x^2 - 7x + 12 = (x - 3)(x - 4) = 0 \] Therefore: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] **Answer:** \( x = 3 \) or \( x = 4 \)

We look for two numbers that multiply to \(-16\) and add up to \(-6\): \[ -8 \times 2 = -16 \quad \text{and} \quad -8 + 2 = -6 \] Factor: \[ x^2 - 6x - 16 = (x - 8)(x + 2) = 0 \] Therefore: \[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Answer: \( x = 8 \) or \( x = -2 \)

\[ x - \frac{6}{x-5}=0 \quad\Longrightarrow\quad x(x-5)-6=0 \] \[ x^2-5x-6=0 \] This quadratic factors: \[ x^2-5x-6=(x-6)(x+1)=0 \] Thus \[ x-6=0\Rightarrow x=6,\qquad x+1=0\Rightarrow x=-1. \] \textbf{Answer: } \(x=6\) or \(x=-1\).

\[ \frac{5x+1}{4}-\frac{2x-1}{2}=x^2 \] Make a common denominator on the left: \[ \frac{5x+1 - 2(2x-1)}{4}=x^2 \quad\Longrightarrow\quad \frac{5x+1-4x+2}{4}=x^2 \] \[ \frac{x+3}{4}=x^2 \quad\Longrightarrow\quad 4x^2-(x+3)=0 \] \[ 4x^2-x-3=0 \] Factor (or use quadratic formula). This factors as: \[ (4x+3)(x-1)=0 \] Thus \[ 4x+3=0\Rightarrow x=-\tfrac{3}{4},\qquad x-1=0\Rightarrow x=1. \] Answers:\(x=1\) or \(x=-\tfrac{3}{4}\).

Factorising: \( x^2 + 5x + 6 = (x + 2)(x + 3) = 0 \). So, the roots are \( x = - 2 \) and \( x = - 3 \).