Question

${6x}^{2} + 7x - 13 = 0$
Sovle by completing the square ​

Answer 1

Answer:

(36x+7x)-13

(43x)-13

x; 43 + 13

x;56

Answer 2

$6x {}^{2} + 7x - 13 = 0 \\ = > ( \sqrt{6} x) {}^{2} + 2 \times \sqrt{6} x \times \frac{7}{2 \sqrt{6} } + ( \frac{7}{2 \sqrt{6} } ) {}^{2} - 13 = ( \frac{7}{2 \sqrt{6} } ) {}^{2} \\ = > ( \sqrt{6} x + \frac{7}{2 \sqrt{6} } ) ^{2} - 13 = \frac{49}{24} \\ = > ( \sqrt{6} x + \frac{7}{2 \sqrt{6} } ) {}^{2} = \frac{49}{24} + 13 \\ = > ( \sqrt{6} x + \frac{7}{2 \sqrt{6} } ) {}^{2} = \frac{49 + 312}{24} \\ = > ( \sqrt{6} x + \frac{7}{2 \sqrt{6} } ) {}^{2} = \frac{361}{24} \\ = > \sqrt{6} x + \frac{7}{2 \sqrt{6} } = ( + - ) \sqrt{ \frac{361}{24} } \\ = > \sqrt{6} x = - \frac{7}{2 \sqrt{6} } ( + - ) \frac{19}{2 \sqrt{6} } \\ = > \sqrt{6} x = \frac{ - 7( + - )19}{2 \sqrt{6} } \\ = > x = \frac{ - 7( + - )19}{2 \sqrt{6} \times \sqrt{6} } \\ = > x = \frac{ - 7( + - )19}{12} \\ = > x = \frac{ - 7 + 19}{12} \: or \: \frac{ - 7 - 19}{12} \\ = > x = \frac{12}{12} \: or \: \frac{ - 26}{12} \\ = > x = 1 \: \: or \: \: \frac{ - 13}{6} \: ans...$

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