Solution:

 \(x^{2} + x - 20 = 0\)

∴ \(x^{2} + x = 20\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }x\right]^{2}\)

= \(\displaystyle \left[\frac {1}{2} \times 1\right]^{2}\)

= \(\displaystyle \left[\frac {1}{2}\right]^{2}\)

= \(\displaystyle \frac {1}{4}\)

∴ Third term = \(\displaystyle \frac {1}{4}\)

Adding \(\displaystyle \frac {1}{4}\) to both sides of equation (i),

 \(\displaystyle x^{2} + x + \frac {1}{4} = 20 + \frac {1}{4}\)

∴ \(\displaystyle x^{2} + x + \frac {1}{4} = \frac {81}{4}\)

∴ \(\displaystyle \left[x + \frac {1}{2}\right]^{2} = \frac {81}{4}\)

Taking square roots of both sides,

 \(\displaystyle x + \frac {1}{2} = \pm \frac {9}{2}\)

∴ \(\displaystyle x = - \frac {1}{2} \pm \frac {9}{2}\)

∴ \(\displaystyle x = - \frac {1}{2} + \frac {9}{2} \text { OR } x = - \frac {1}{2} - \frac {9}{2}\)

∴ \(\displaystyle x = \frac {- 1 + 9}{2} \text { OR } x = \frac {- 1 - 9}{2}\)

∴ \(\displaystyle x = \frac {\cancelto {4}{8}}{\cancelto {1}{2}} \text { OR } x = \frac {\cancelto {- 5}{- 10}}{\cancelto {1}{2}}\)

∴ \(\displaystyle x = 4 \text { OR } x = - 5\)

∴ 4, − 5 are the roots of the given quadratic equation.

Solution:

 \(x^{2} + 2x - 5 = 0\)

∴ \(x^{2} + 2x = 5\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }x\right]^{2}\)

= \(\displaystyle \left[\frac {1}{\cancel {2}} \times \cancel {2}\right]^{2}\)

= \(1^{2}\)

= 1

∴ Third term = 1

Adding 1 to both sides of equation (i),

 \(\displaystyle x^{2} + 2x + 1 = 5 + 1\)

∴ \(\displaystyle x^{2} + x + 1 = 6\)

∴ \(\displaystyle (x + 1)^{2} = 6\)

Taking square roots of both sides,

 \(\displaystyle x + 1 = \pm \sqrt {6}\)

∴ \(\displaystyle x = - 1 \pm \sqrt {6}\)

∴ \(\displaystyle x = - 1 + \sqrt {6} \text { OR } x = - 1 - \sqrt {6}\)

∴ \(\displaystyle - 1 + \sqrt {6}, - 1 - \sqrt {6}\) are the roots of the given quadratic equation.

Solution:

 \(m^{2} - 5m = - 3\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }m\right]^{2}\)

= \(\displaystyle \left[\frac {1}{2} \times {- 5}\right]^{2}\)

= \(\displaystyle \left[\frac {- 5}{2}\right]^{2}\)

= \(\displaystyle \frac {25}{4}\)

∴ Third term = \(\displaystyle \frac {25}{4}\)

Adding \(\displaystyle \frac {25}{4}\) to both sides of equation (i),

 \(\displaystyle m^{2} - 5m + \frac {25}{4} = - 3 + \frac {25}{4}\)

∴ \(\displaystyle m^{2} - 5m + \frac {25}{4} = \frac {- 12 + 25}{4}\)

∴ \(\displaystyle m^{2} - 5m + \frac {25}{4} = \frac {13}{4}\)

Taking square roots of both sides,

 \(\displaystyle m - \frac {5}{2} = \pm \sqrt {\frac {13}{4}}\)

∴ \(\displaystyle m - \frac {5}{2} = \pm \frac{\sqrt {13}}{2}\)

∴ \(\displaystyle m = \frac {5}{2} \pm \frac {\sqrt {13}}{2}\)

∴ \(\displaystyle m = \frac {5}{2} + \frac {\sqrt {13}}{2} \text { OR } m = \frac {5}{2} - \frac {\sqrt {13}}{2}\)

∴ \(\displaystyle m = \frac {5 + \sqrt {13}}{2} \text { OR } m = \frac {5 - \sqrt {13}}{2}\)

∴ \(\displaystyle \frac {5 + \sqrt {13}}{2}, \frac {5 - \sqrt {13}}{2}\) are the roots of the given quadratic equation.

Solution:

 \(9y^{2} - 12y + 2 = 0\)

∴ \(9y^{2} - 12y = - 2\)

Dividing both sides by 9,

 \(\displaystyle \frac {\bcancel {9}y^{2}}{\bcancel {9}} - \frac {\cancelto {4}{12}y}{\cancelto {3}{9}} = \frac {- 2}{9}\)

∴ \(\displaystyle y^{2} - \frac {4}{3}y = - \frac {2}{9}\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }y\right]^{2}\)

= \(\displaystyle \left[\frac {1}{\cancelto {1}{2}} \times - \frac{\cancelto {2}{4}}{3}\right]^{2}\)

= \(\displaystyle \left[- \frac{2}{3}\right]^{2}\)

= \(\displaystyle \frac{4}{9}\)

∴ Third term = \(\displaystyle \frac {4}{9}\)

Adding \(\displaystyle \frac {4}{9}\) to both sides of equation (i),

 \(\displaystyle y^{2} - \frac{4}{3}y + \frac{4}{9} = - \frac{2}{9} + \frac{4}{9}\)

∴ \(\displaystyle \left[y - \frac{2}{3}\right]^{2} = \frac{- 2 + 4}{9}\)

∴ \(\displaystyle \left[y - \frac{2}{3}\right]^{2} = \frac{2}{9}\)

Taking square roots of both sides,

 \(\displaystyle y - \frac{2}{3} = \pm \sqrt {\frac{2}{9}}\)

 \(\displaystyle y - \frac{2}{3} = \pm {\frac{\sqrt {2}}{3}}\)

 \(\displaystyle y = \frac{2}{3} \pm {\frac{\sqrt {2}}{3}}\)

 \(\displaystyle y = \frac{2}{3} + {\frac{\sqrt {2}}{3}} \text { OR } y = \frac{2}{3} - {\frac{\sqrt {2}}{3}}\)

∴ \(\displaystyle y = \frac{2 + \sqrt {2}}{3} \text { OR } y = \frac{2 - \sqrt {2}}{3}\)

∴ \(\displaystyle \frac{2 + \sqrt {2}}{3}, \frac{2 - \sqrt {2}}{3}\) are the roots of the given quadratic equation.

Solution:

 \(2y^{2} + 9y + 10 = 0\)

∴ \(2y^{2} + 9y = - 10\)

Dividing both sides by 2,

 \(\displaystyle \frac {\bcancel {2}y^{2}}{\bcancel {2}} + \frac {9y}{2} = \frac {\cancelto {- 5}{- 10}}{\cancelto {1}{2}}\)

∴ \(\displaystyle y^{2} + \frac {9}{2}y = - 5\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }y\right]^{2}\)

= \(\displaystyle \left[\frac {1}{2} \times \frac{9}{2}\right]^{2}\)

= \(\displaystyle \left[\frac{9}{4}\right]^{2}\)

= \(\displaystyle \frac{81}{16}\)

∴ Third term = \(\displaystyle \frac {81}{16}\)

Adding \(\displaystyle \frac {81}{16}\) to both sides of equation (i),

 \(\displaystyle y^{2} + \frac{9}{2}y + \frac{81}{16} = - 5 + \frac{81}{16}\)

∴ \(\displaystyle \left[y + \frac{9}{4}\right]^{2} = \frac{- 80 + 81}{16}\)

∴ \(\displaystyle \left[y + \frac{9}{4}\right]^{2} = \frac{1}{16}\)

Taking square roots of both sides,

 \(\displaystyle y + \frac{9}{4} = \pm \sqrt {\frac{1}{16}}\)

∴ \(\displaystyle y + \frac{9}{4} = \pm {\frac{1}{4}}\)

∴ \(\displaystyle y = - \frac{9}{4} \pm {\frac{1}{4}}\)

∴ \(\displaystyle y = - \frac{9}{4} + {\frac{1}{4}} \text { OR } y = - \frac{9}{4} - {\frac{1}{4}}\)

∴ \(\displaystyle y = \frac{- 9 + 1}{4} \text { OR } y = \frac{- 9 - 1}{4}\)

∴ \(\displaystyle y = \frac{\cancelto {- 2}{- 8}}{\cancelto {1}{4}} \text { OR } y = \frac{\cancelto {- 5}{- 10}}{\cancelto {2}{4}}\)

∴ \(\displaystyle y = - 2 \text { OR } y = - \frac{5}{2}\)

∴ \(\displaystyle - 2, - \frac{5}{2}\) are the roots of the given quadratic equation.

Solution:

 \(5x^{2} = 4x + 7\)

∴ \(5x^{2} - 4x = 7\)

Dividing both sides by 5,

 \(\displaystyle \frac {\bcancel {5}x^{2}}{\bcancel {5}} - \frac {4x}{5} = \frac {7}{5}\)

∴ \(\displaystyle x^{2} - \frac {4}{5}x = \frac {7}{5}\) ... (i)

Now, third term

= \(\displaystyle \left[\frac {1}{2} \times \text { Coefficient of }x\right]^{2}\)

= \(\displaystyle \left[\frac {1}{\cancelto {1}{2}} \times - \frac{\cancelto {2}{4}}{5}\right]^{2}\)

= \(\displaystyle \left[- \frac{2}{5}\right]^{2}\)

= \(\displaystyle \frac{4}{25}\)

∴ Third term = \(\displaystyle \frac {4}{25}\)

Adding \(\displaystyle \frac {4}{25}\) to both sides of equation (i),

 \(\displaystyle x^{2} - \frac{4}{5}x + \frac{4}{25} = \frac{7}{5} + \frac{4}{25}\)

∴\(\displaystyle x^{2} - \frac{4}{5}x + \frac{4}{25} = \frac{35}{25} + \frac{4}{25}\)

∴ \(\displaystyle \left[x - \frac{2}{5}\right]^{2} = \frac{35 + 4}{25}\)

∴ \(\displaystyle \left[x - \frac{2}{5}\right]^{2} = \frac{39}{25}\)

Taking square roots of both sides,

 \(\displaystyle x - \frac{2}{5} = \pm \sqrt {\frac{39}{25}}\)

∴ \(\displaystyle x - \frac{2}{5} = \pm {\frac{\sqrt {39}}{5}}\)

∴ \(\displaystyle x = \frac{2}{5} \pm {\frac{\sqrt {39}}{5}}\)

∴ \(\displaystyle x = \frac{2}{5} + {\frac{\sqrt {39}}{5}} \text { OR } x = \frac{2}{5} - {\frac{\sqrt {39}}{5}}\)

∴ \(\displaystyle x = \frac{2 + \sqrt {39}}{5} \text { OR } x = \frac{2 - \sqrt {39}}{5}\)

∴ \(\displaystyle \frac{2 + \sqrt {39}}{5}, \frac{2 - \sqrt {39}}{5}\) are the roots of the given quadratic equation.