Answer:
[tex]x=\dfrac{-3\pm3\sqrt{5}}{2}[/tex]
Step-by-step explanation:
Given equation:
[tex]x^2+3x-9=0[/tex]
Completing the square
Move the constant to the right side by adding 9 to both sides:
[tex]\implies x^2+3x=9[/tex]
Add the square of half the coefficient of x to both sides:
[tex]\implies x^2+3x+\left(\dfrac{3}{2}\right)^2=9+\left(\dfrac{3}{2}\right)^2[/tex]
[tex]\implies x^2+3x+\dfrac{9}{4}=\dfrac{45}{4}[/tex]
Factor the trinomial on the left side:
[tex]\implies \left(x+\dfrac{3}{2}\right)^2=\dfrac{45}{4}[/tex]
Square root both sides:
[tex]\implies x+\dfrac{3}{2}=\pm\sqrt{\dfrac{45}{4}}[/tex]
[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{45}}{\sqrt{4}}}[/tex]
[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{9 \cdot 5}}{2}[/tex]
[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{9} \sqrt{5}}{2}[/tex]
[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{3\sqrt{5}}{2}[/tex]
Subtract 3/2 from both sides:
[tex]\implies x=\pm\dfrac{3\sqrt{5}}{2}-\dfrac{3}{2}[/tex]
[tex]\implies x=\dfrac{-3\pm3\sqrt{5}}{2}[/tex]
