QUESTION 1 Solve for \( x \) \( \begin{array}{l}1.1 \quad x^{2}+4 x-21=0 \\ 1.2 \quad 3 x^{2}-5 x=1 \\ 1.3 \quad \sqrt{2-x}=x-2 \\ \text { QUESTION } 2 \\ 2.1 \quad x^{2}-4 x+2=0 \\ 2.2 \text { Solve for } x \text { by completing the square: } \\ \text { Given the following the equation } f(x)=5 x^{2}-12 x+4 \\ \text { Solve for } x \text { if } f(x)>0 \\ 2.3 \text { If } \frac{6}{\sqrt{2}-2}=m+n \sqrt{t} \\ \text { Determine the values of } m, n \text { and } t\end{array} \)

To solve \( x^{2}+4x-21=0 \), we can factor it as \( (x+7)(x-3)=0 \). Thus, the solutions are \( x = -7 \) and \( x = 3 \). For \( 3x^2-5x=1 \), rearranging gives \( 3x^2-5x-1=0 \). Using the quadratic formula, we find \( x = \frac{5 \pm \sqrt{25 + 12}}{6} = \frac{5 \pm \sqrt{37}}{6} \). This gives two solutions: \( x = \frac{5 + \sqrt{37}}{6} \) and \( x = \frac{5 - \sqrt{37}}{6} \). For \( \sqrt{2-x}=x-2 \), squaring both sides leads to the equation \( 2-x=(x-2)^2 \). Solving this results in \( x = \frac{2+\sqrt{10}}{2} \) and \( x = \frac{2-\sqrt{10}}{2} \) after simplifying. As for \( x^{2}-4x+2=0 \), applying the quadratic formula reveals the solutions to be \( x = 2 \pm \sqrt{2} \). In the completion of the square for \( f(x)=5x^{2}-12x+4 \), rewriting it reveals that \( f(x) > 0 \) when the resulting quadratic is greater than zero, leading to determining values at which the function is positive. Lastly, to simplify \( \frac{6}{\sqrt{2}-2} \), we multiply the numerator and denominator by \( \sqrt{2}+2 \). This yields \( m = 3(\sqrt{2}+2) \), \( n = 3 \), and \( t=2 \). For the complete analysis and exploration of how quadratics affect various real-world situations, one might want to check out some math-focused applications like projectile motion or optimization problems. If you're interested in diving deeper, I recommend exploring resources that break down the quadratic formula step-by-step or watching videos that visualize graphs of quadratic equations. They can really help solidify your understanding!