(i) \(2 x^{2}-5 x+2=0\)
The formula for finding roots of a quadratic equation \(a x^{2}+b x+c=0\) is
$$
\begin{aligned}
&x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&2 x^{2}-5 x+2=0 \\
&a b c=\frac{-(-5) \pm \sqrt{(-5)^{2}-4 \times 2 \times 2}}{2 \times 2}
\end{aligned}
$$
\(=\frac{5 \pm \sqrt{25-16}}{4}\)
\(=\frac{5 \pm \sqrt{9}}{4}=\frac{5 \pm 3}{4}=\)
\(=2, \frac{1}{2}\)
\(\therefore\) Solutions is \(2, \frac{1}{2}\)
(ii) \(\sqrt{2} \mathrm{f}^{2}-6 \mathrm{f}+3 \sqrt{2}\)
\(\sqrt{2} f^{2}-6 f+3 \sqrt{2}=0\)
\(a \quad b \quad c\)
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here \(f=\frac{-(-6) \pm \sqrt{(-6)^{2}-4 \times \sqrt{2} \times 3 \sqrt{2}}}{2 \times \sqrt{2}}\)
\(-\frac{6 \pm \sqrt{36-24}}{2 \sqrt{2}}\)
\(=\frac{6 \pm \sqrt{12}}{2 \sqrt{2}}=\frac{6 \pm 2 \sqrt{3}}{2 \sqrt{2}}=\frac{2(3 \pm \sqrt{3})}{22 \sqrt{2}}\)
\(\Rightarrow \frac{3+\sqrt{3}}{\sqrt{2}}, \frac{3-\sqrt{3}}{2}\)
(iii) \(3 y^{2}-20 y-3=0\)
$$
\begin{aligned}
&\begin{array}{l}
3 y^{2}+20 y-23=0 \\
a \quad b \quad c
\end{array} \\
&x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&\text { Here } y=\frac{-(-20) \pm \sqrt{(-20)^{2}-4 \times 3 \times-23}}{2 \times 3} \\
&=\frac{20 \pm \sqrt{400+276}}{6} \\
&=\frac{20 \pm \sqrt{676}}{6}=\frac{20 \pm 26}{6} \\
&=\frac{46}{6} \text { or } \frac{-6}{6} \\
&y \Rightarrow \frac{23}{3} \text { or }-1
\end{aligned}
$$
$$
\begin{aligned}
&\text { (iv) } 36 y^{2}-12 a y+\left(a^{2}-b^{2}\right)=0 \\
&\begin{array}{l}
36 y^{2}-12 a y+\left(a^{2}-b^{2}\right)=0 \\
a \\
b
\end{array} \\
&x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\
&\text { Here } y=\frac{-(-12 a) \pm \sqrt{(-12 a)^{2}-4 \times 36 \times\left(a^{2}-b^{2}\right)}}{2 \times 36} \\
&=\frac{12 a \pm \sqrt{144 a^{2}-144 a^{2}+144 b^{2}}}{72} \\
&=\frac{12 a \pm 12 b}{72} \Rightarrow \frac{a \pm b}{6}
\end{aligned}
$$
