Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.
Solution:
Given equation,
$2{{x}^{2}}-7x+3=0$
$2\left( {{x}^{2}}-\frac{7x}{2}+\frac{3}{2} \right)=0$
${{x}^{2}}-2\times \frac{1}{2}\times \frac{7}{2}\times x+\frac{3}{2}=0$
${{x}^{2}}-2\times \frac{7}{4}\times x+{{\left( \frac{7}{4} \right)}^{2}}-{{\left( \frac{7}{4} \right)}^{2}}+\frac{3}{2}=0$
${{\left( x-\frac{7}{4} \right)}^{2}}-\frac{49}{16}+\frac{3}{2}=0$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49}{16}-\frac{3}{2}$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49-26}{16}$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{25}{16}$
${{\left( x-\frac{7}{4} \right)}^{2}}={{\left( \frac{5}{4} \right)}^{2}}$
$x-\frac{7}{4}=\pm \frac{5}{4}$
$x-\frac{7}{4}=\frac{5}{4}orx-\frac{7}{4}=-\frac{5}{4}$
$x=\frac{7}{4}+\frac{5}{4}orx=\frac{7}{4}-\frac{5}{4}$
$2\left( {{x}^{2}}-\frac{7x}{2}+\frac{3}{2} \right)=0$
${{x}^{2}}-2\times \frac{7}{2}\times \frac{1}{2}\times x+\frac{3}{2}=0$
${{x}^{2}}-2\times \frac{7}{2}\times x+{{\left( \frac{7}{4} \right)}^{2}}-{{\left( \frac{7}{4} \right)}^{2}}+\frac{3}{2}=0$
${{x}^{2}}-2\times \frac{7}{4}\times x+{{\left( \frac{7}{4} \right)}^{2}}-\left( \frac{49}{16} \right)+\frac{3}{2}=0$
${{\left( x-\frac{7}{4} \right)}^{2}}-\frac{49}{16}+\frac{3}{2}=0$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49}{16}-\frac{3}{2}$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49-26}{16}$
${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{25}{16}$
${{\left( x-\frac{7}{4} \right)}^{2}}={{\left( \frac{5}{4} \right)}^{2}}$
$x-\frac{7}{4}=\pm \frac{5}{4}$
$x-\frac{7}{4}=\frac{5}{4}orx-\frac{7}{4}=-\frac{5}{4}$
$x=\frac{7}{4}+\frac{5}{4}orx=\frac{7}{4}-\frac{5}{4}$
$\Rightarrow x=12/4=3orx=2/4=1/2$
Thus, the roots of the given quadratic equation are $x=3$ and $x=1/2.$