Question

find the roots of the equation

Answer 1

if we find x
$x + \sqrt{x + 2 } = 4$
on square on both side form we get
${x}^{2} + x + 2 = 16$
now we get quadratic equation which solve by 3 method , but simplest one by quadratic equation.

${x}^{2} + x - 14= 0$
where b is 2nd no , a is 1 and and c is 3rd.

$x = \frac{ - b + \sqrt{b ^{2} + 4ab } }{2a}$
$x = \frac{ - b - \sqrt{ {b}^{2} + 4ac} }{2a}$

there is always two ans of x one with + sign after -b and other is - sign after -b.

the ans is
$x = \frac{ - 1 + \sqrt{57} }{2}$
$x = \frac{ - 1 - \sqrt{57} }{2}$