Answer:
[tex]\large {\textsf{B and E}}\ \implies x_1=-2\sqrt{5}-3,\ x_2=+2\sqrt{5}-3[/tex]
Step-by-step explanation:
Quadratic Formula: [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Quadratic Equation: ax² + bx + x = 0, where a ≠ 0
Given equation: x² + 6x + 9 = 20
Step 1: Subtract 20 from both sides.
x² + 6x + 9 - 20 = 20 - 20
⇒ x² + 6x - 11 = 0
Step 2: Identify the values of a, b, and c and substitute them in the formula.
⇒ a = 1, b = 6, c = -11
[tex]x=\dfrac{-6\pm\sqrt{\bold{6^2}-4(1)(-11)}}{\bold{2(1)}}\\\\x=\dfrac{-6\pm\sqrt{36\bold{\ - \ 4(-11)}}}{2}\\\\x=\dfrac{-6\pm\sqrt{\bold{36+44}}}{2}\\\\x=\dfrac{-6\pm\sqrt{80}}{2}\\\\x=\dfrac{-6\pm\sqrt{16\times5}}{2}\implies \dfrac{-6\pm\sqrt{\bold{16}}\times\sqrt{5}}{2} \\\\x=\dfrac{-6\pm4\sqrt{5}}{2}[/tex]
Step 3: Separate into possible cases.
[tex]x_1=\dfrac{-6-4\sqrt{5}}{2}\implies \dfrac{-6}{2}+\dfrac{-4\sqrt{5}}{2}\implies \boxed{-3-2\sqrt{5}\ \ \textsf{or}\ -2\sqrt{5}-3}\\\\x_2=\dfrac{-6+4\sqrt{5}}{2}\implies \dfrac{-6}{2}+\dfrac{4\sqrt{5}}{2}\implies \boxed{-3+2\sqrt{5}\ \ \textsf{or}\ \ 2\sqrt{5}-3}[/tex]
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