Question

$4x ^{2} + 3x + 5 = 10$
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Answer 1

$= = > \: \: 4 {x}^{2} + 3x + 5 = 10 \\ \\ = = > \: \: 4 {x}^{2} + 3x = 5 \\ \\ = = > \: \: {(2x)}^{2} + 2 \times 2x \times \frac{3}{4} = 5 \\ \\ = = > \: \: {(2x)}^{2} + 2 \times 2x \times \frac{3}{4} + { \left( \frac{3}{4} \right)}^{2} = 5 + { \left( \frac{3}{4} \right)}^{2} \\ \\ = = > \: \: { \left( 2x + \frac{3}{4} \right)}^{2} = 5 + \frac{9}{16} \\ \\ = = > \: \: { \left( 2x + \frac{3}{4} \right)}^{2} = \frac{89}{16} \\ \\ = = > \: \: 2x + \frac{3}{4} = \pm \: \frac{ \sqrt{89} }{4}$

Case - +ve

$= = > \: \: 2x + \frac{3}{4} = \frac{ \sqrt{89} }{4} \\ \\ = = > \: \: 2x = \frac{ \sqrt{89} }{4} - \frac{3}{4} \\ \\ = = > \: \: 2x = \frac{ \sqrt{89} - 3}{4} \\ \\ = = > \: \: x = \frac{ \sqrt{89} - 3 }{8}$

Case - (-ve)

$= = > \: \: 2x + \frac{3}{4} = - \frac{ \sqrt{89} }{4} \\ \\ = = > \: \: 2x = - \frac{ \sqrt{89}}{4} - \frac{3}{4} \\ \\ = = > \: \: x = \frac{ - \sqrt{89} - 3}{8} \\ \\ = = > \: \: x = \frac{ - ( \sqrt{89} + 3)}{8}$