Question

please solve this sum by step by step​ and i will mark as brainlist​

Answer 1

hope it helped u!

...................

Answer 2

$\\ \red{ \bold{ \underline{ \bf{ANSWER }}}}$

$given \\ a \: {x}^{2} + b \: x + c = 0 \\ where \: \: \: \: \: \: a \neq 0$

because it is not then equation become linear.

$now \\ a \: {x}^{2} + b \: x + c =0 \\ \: {x}^{2} + \frac{b}{a} \: x = - \frac{c}{a} \\ \: {x}^{2} + \frac{b}{a} \: x + { (\frac{b}{2a}) }^{2} = - \frac{c}{a} \: + \: {(\frac{b}{2a} )}^{2} \\ (x + \frac{b}{2a} )^2 = - \frac{c}{a} \: + \: {(\frac{b}{2a} )}^{2} \\ (x + \frac{b}{2a} ) = \pm \: \sqrt{ - \frac{c}{a} \: + \: {(\frac{b}{2a} )}^{2}}$

$\\ (x ) = - \frac{b}{2a} \pm \:\sqrt{ - \frac{c}{a} \: + \: {(\frac{b}{2a} )}^{2}}$

take lcm

and solve

we get

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

cheers