Question

please solve it ok............​

Answer 1

$\sf \: {ax}^{2} + bx + c = 0 \\ \\ \\ \sf \: x = \frac{b \: ± \: \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \\ \sf \: x = \frac{ - 7± \sqrt{ {7}^{2} - 4( \sqrt{2} )(5 \sqrt{2}) } }{2 \sqrt{2} } \\ \\ \\ \sf \: x = \frac{ - 7± \sqrt{49 - 40} }{2 \sqrt{2} } \\ \\ \\ \sf \implies \: x = \frac{ - 7±3}{2 \sqrt{2} } \\ \\ \\ \sf \: x_{1} = \frac{ - 4}{2 \sqrt{2} } = - \sqrt{2} \\ \\ \\ \sf \: x_{2} = \frac{ - 10}{2 \sqrt{2} } = \frac{ - 5}{ \sqrt{2} }$

Answer 2

Answer:

$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$

$=> \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2} = 0$

$=> \sqrt{2} x(x + \sqrt{2} ) + 5(x + \sqrt{2} ) = 0$

$=> (x + \sqrt{2} )( \sqrt{2} x + 5) = 0$

$if \: \: x + \sqrt{2} = 0 \\ => x = - \sqrt{2}$

$or \: \: \sqrt{2} x + 5 = 0 \\ => \sqrt{2} x = - 5 \\ => x = - \frac{5}{ \sqrt{2} } = - \frac{5 \sqrt{2} }{2}$

Explanation:

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