Question

x^2+6x-4=0quadratic equation​

Answer 1

Step-by-step explanation:

the answer is in the attachment to

Answer 2

SOLUTION :

$\star \: \: \sf{\underline{Given : }}$

$\sf {x}^{2} + 6x - 4 = 0$

Process 1

$\sf {x}^{2} + 6x - 4 = 0$

$\sf \implies {x}^{2} +6x + 9- 9 - 4 = 0$

$\sf \implies {(x + 3)}^{2} - 13 = 0$

$\sf \implies {(x + 3)}^{2} - { \sqrt{13} }^{2} = 0$

$\sf \implies \: (x + 3 + \sqrt{13} )(x + 3 - \sqrt{13} ) = 0$

$\sf \implies { \boxed {\sf{ x = - (3 + \sqrt{13} ) \: \: \: or \: \: \: x = (\sqrt{13} - 3)}}}$

Process 2

$\sf {x}^{2} + 6x - 4 = 0$

We know quadratic equation formula

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

Putting value,

$\implies \sf \: x = \frac{ - 6± \sqrt{ {6}^{2} - 4 \times ( - 4) \times 1} }{2 \times 1}$

$\implies { \sf \: x = - 3 + \frac{ \sqrt{52} }{2} \: \: \: or \: \: \: x = - 3 - \frac{ \sqrt{52} }{2} }$

$\implies \boxed{ \sf \: x = - 3 + \sqrt{13} \: \: \: or \: \: \: x = - 3 - \sqrt{13} }$