Answer:
[tex]\boxed {x = \sqrt{3} - 3}[/tex]
[tex]\boxed {x = -\sqrt{3} - 3}[/tex]
Step-by-step explanation:
Solve for the value of [tex]x[/tex]:
[tex]x^2 + 6x + 6 = 0[/tex]
-When you use the quadratic formula ( [tex]\frac{-b\pm \sqrt{b^{2} - 4ac}}{2a}[/tex] ), it would give you two solutions. So, use the quadratic formula:
[tex]x = \frac{-6\pm \sqrt{6^{2} - 4 \times 6}}{2}[/tex]
-Simplify [tex]6[/tex] by the exponent [tex]2[/tex]:
[tex]x = \frac{-6\pm \sqrt{6^{2} - 4 \times 6}}{2}[/tex]
[tex]x = \frac{-6\pm \sqrt{36 - 4 \times 6}}{2}[/tex]
-Multiply both [tex]-4[/tex] and [tex]6[/tex]:
[tex]x = \frac{-6\pm \sqrt{36 - 4 \times 6}}{2}[/tex]
[tex]x = \frac{-6\pm \sqrt{36 - 24}}{2}[/tex]
-Add [tex]34[/tex] and [tex]-24[/tex]:
[tex]x = \frac{-6\pm \sqrt{36 - 24}}{2}[/tex]
[tex]x = \frac{-6\pm \sqrt{12}}{2}[/tex]
-Take the square root of [tex]12[/tex]:
[tex]x = \frac{-6\pm \sqrt{12}}{2}[/tex]
[tex]x = \frac{-6\pm 2\sqrt{3}}{2}[/tex]
-Now solve the equation when [tex]\pm[/tex] is plus, So, add [tex]-6[/tex] to [tex]2\sqrt{3}[/tex]:
[tex]x = \frac{-6\pm 2\sqrt{3}}{2}[/tex]
[tex]x = \frac{2\sqrt{3} - 6}{2}[/tex]
-Divide [tex]-6 + 2\sqrt{3}[/tex] both sides by [tex]2[/tex]:
[tex]x = \frac{2\sqrt{3} - 6}{2}[/tex]
[tex]\boxed {x = \sqrt{3} - 3}[/tex] (Answer 1)
-Now solve the equation when [tex]\pm[/tex] is minus. So, Subtract [tex]2\sqrt{3}[/tex] from [tex]-6[/tex]:
[tex]x = \frac{-2\sqrt{3} - 6}{2}[/tex]
-Divide [tex]-2\sqrt{3} - 6[/tex] by [tex]2[/tex]:
[tex]x = \frac{-2\sqrt{3} - 6}{2}[/tex]
[tex]\boxed {x = -\sqrt{3} - 3}[/tex] (Answer 2)
