Question

root 2x square +7x +5root 2=0 by quadratic formula​

Answer 1

$\boxed{SOLUTION \: by \: RKRAUSHAN}$

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

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

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

$= > x = - \sqrt{2} \\ \\ = > x = \frac{ - 5}{ \sqrt{2} } = > \frac{ - 5 \sqrt{2} }{2} \\ \\ on \: rationalising \:$

Or

By Quadratic Method

In Attachment

$\huge \underline \red {Hope \: it \: helps} \\ \huge \underline \green{Thank \: U}$