Question

10x2+5x-2=0 by quadraric method
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Answer 1

I think your question is wrong it can't be solved by quadratic method but can be solved by other methods.

Answer 2

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$\: \: \bigstar \large \green{ \bold{ \underline{ \underline{ \: correct \: answer}}}}$

$\: \large \underline{\boxed{ \bf{ \color{cyan}given}}} :$

$\: \bf{10 {x}^{2} + 5x - 2 = 0 \: by \: quadratic \: equation}$

$\: \: \large \orange{ \bold{ \underline{ \underline{correct \: explaination}}}}$

$\: \implies \displaystyle \sf{10 {x}^{2} + 5x - 2 = 0}$

$\: \: \implies \underline{ \diamondsuit \: \displaystyle \sf{by \: determinant \: form}}$

$\: \: \implies \displaystyle \sf{ {b}^{2} - 4ac}$

$\: \: \implies \displaystyle \sf{ {(5)}^{2} - 4(10) ( - 2)}$

$\: \: \implies \displaystyle \sf{25 + 80} = 105$

$\: \: \\ \implies \underline{\displaystyle \sf{ \diamondsuit \: by \: formula \:method}}$

$\: \: \implies \displaystyle \sf{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

$\: \: \implies \displaystyle \sf{x = \frac{ - 5 \pm \sqrt{105} }{2 \times 10} }$

$\: \: \: \\ \implies \displaystyle \sf{x = \frac{ - 5 \pm \sqrt{105}}{20} }$

$\: \: \\ \implies \displaystyle \sf{x = \frac{ - 5 + \sqrt{105} }{20} \: and \: x = \frac{ - 5 - \sqrt{105} }{20} }$

$\: \heartsuit \boxed{ \boxed{ \underline{ \bf{the \: roots \: are \: x = \frac{ - 5 - \sqrt{105} }{20} \: and \: x = \frac{ - 5 + \sqrt{105} }{20} }}}}$