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(D) Given equation: $3y^{2} + 7y - 20 = 0$.Divide by $3$: $y^{2} + \frac{7}{3}y - \frac{20}{3} = 0$.Add and subtract $(\frac{1}{2} 	imes 	ext{coeffi

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$-4, \frac{5}{3}$

Vedclass · Answer

(D) Given equation: $3y^{2} + 7y - 20 = 0$.Divide by $3$: $y^{2} + \frac{7}{3}y - \frac{20}{3} = 0$.Add and subtract $(\frac{1}{2} 	imes 	ext{coefficient of } y)^{2} = (\frac{7}{6})^{2} = \frac{49}{36}$:$y^{2} + \frac{7}{3}y + \frac{49}{36} - \frac{49}{36} - \frac{20}{3} = 0$.$(y + \frac{7}{6})^{2} - (\frac{49 + 240}{36}) = 0$.$(y + \frac{7}{6})^{2} = \frac{289}{36}$.Taking square root: $y + \frac{7}{6} = \pm \frac{17}{6}$.Case $1$: $y = \frac{17}{6} - \frac{7}{6} = \frac{10}{6} = \frac{5}{3}$.Case $2$: $y = -\frac{17}{6} - \frac{7}{6} = -\frac{24}{6} = -4$.Thus,the roots are $-4, \frac{5}{3}$.

Obtain the roots of the following equation using the method of 'completing the square': $3y^{2} + 7y - 20 = 0$.

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