Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Given equation,

$2{{x}^{2}}+x-4=0$

$2\left( {{x}^{2}}+\frac{x}{2}-\frac{4}{2} \right)=0$

${{x}^{2}}+2\times \frac{1}{2}\times \frac{1}{2}\times x-2=0$

${{x}^{2}}+2\times \frac{1}{4}\times x+{{\left( \frac{1}{4} \right)}^{2}}-{{\left( \frac{1}{4} \right)}^{2}}-2=0$

${{\left( x+\frac{1}{4} \right)}^{2}}={{\left( \frac{1}{4} \right)}^{2}}+2$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1}{16}+2$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1+2\times 16}{16}$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1+32}{16}$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{33}{16}$

$\left( X+\frac{1}{4} \right)=\pm \sqrt{\frac{33}{16}}$

$\left( x+\frac{1}{4} \right)=\sqrt{\frac{33}{16}}$

or $\left( x+\frac{1}{4} \right)=-\sqrt{\frac{33}{16}}$

$x=\frac{\sqrt{33}}{4}-\frac{1}{4}orx=-\frac{\sqrt{33}}{4}-\frac{1}{4}$

$x=\frac{\sqrt{33}-1}{4}orx=\frac{\sqrt{33}-1}{4}$

$x=\frac{\sqrt{33}-1}{4}orx=\frac{-\sqrt{33}-1}{4}$

$2\left( {{x}^{2}}+\frac{x}{2}-\frac{4}{2} \right)=0$

${{x}^{2}}+2\times \frac{1}{2}\times \frac{1}{2}\times x-2=0$

${{x}^{2}}+2\times \frac{1}{4}\times x+{{\left( \frac{1}{4} \right)}^{2}}-{{\left( \frac{1}{4} \right)}^{2}}-2=0$

${{\left( x+\frac{1}{4} \right)}^{2}}={{\left( \frac{1}{4} \right)}^{2}}+2$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1}{16}+2$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1+2\times 16}{16}$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1+32}{16}$

${{\left( x+\frac{1}{4} \right)}^{2}}=\frac{33}{16}$

$\left( x+\frac{1}{4} \right)=\pm \sqrt{\frac{33}{16}}$

$\left( x+\frac{1}{4} \right)=\sqrt{\frac{33}{16}}$

or $\left( x+\frac{1}{4} \right)=-\sqrt{\frac{33}{16}}$

$x=\frac{\sqrt{33}}{4}-\frac{1}{4}orx=-\frac{\sqrt{33}}{4}-\frac{1}{4}$

$x=\frac{\sqrt{33}-1}{4}orx=-\frac{\sqrt{33}-1}{4}$

Thus, the roots of the given quadratic equation are