Question

x^2+6x+6=0 solve using quadratic formula​

Answer 1

Answer:

$\mathfrak{\pink{The\:roots\:of\:the\:given\:equation\:are\:x = - 3 + \sqrt{3}\:and\:x = - 3 - \sqrt{3}}}$

Step-by-step explanation:

x^2 + 6x + 6 = 0

Here, a = 1, b = 6 and c = 6

$\mathfrak{\purple{By\:using\:quadratic\:formula,}}$

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

$x = \frac{-(6) \pm \sqrt{(6)^2 - 4(1)(6)}}{2(1)}$

$x = \frac{-6 \pm \sqrt{36 - 24}}{2}$

$x = \frac{-6 \pm \sqrt{12}}{2}$

$x = \frac{-6 \pm \sqrt{4 \times 3}}{2}$

$x = \frac{-6 \pm 2\sqrt{3}}{2}$

$x = \frac{2(-3 \pm \sqrt{3})}{2}$

$x = \frac{\cancel{2}(-3 \pm \sqrt{3})}{\cancel{2}}$

$x = -3 \pm \sqrt{3}$

$x = - 3 + \sqrt{3}\:or\:x = - 3 - \sqrt{3}$

Therefore, the roots of the given equation are $x = - 3 + \sqrt{3}\:and\:x = - 3 - \sqrt{3}$

Answer 2

Step-by-step explanation:

$hello... \\ {x}^{2} + 6x + 6 = 0 \\ a = 1 \\ b = 6 \\ c = 6 \\ \\ by \: quadratic \: formula = > \\ \\ \frac{ - b + \: or \: - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \frac{ - 6 + \: or \: - \sqrt{36 - 24} }{2(1)} \\ \frac{ - 6 + \: or \: - 4 \sqrt{ 3} }{2} \\ \frac{ - 3 + \: or \: - 2 \sqrt{3} }{1} \\ \\ - 3 + 2 \sqrt{3} \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: - 3 - 2 \sqrt{3} \\ \\ hope \: it \: helps \: uh..$