Answer:
[tex]x=\frac{-1+\sqrt{5}i}{3}[/tex] or [tex]x=\frac{-\sqrt{5}i-1}{3}[/tex]
Step-by-step explanation:
The given equation is
[tex]-\frac{3}{2}x^2=x+1[/tex]
We multiply through by -2 to get;
[tex]3x^2=-2x-2[/tex]
We rewrite in standard quadratic equation form to obtain;
[tex]3x^2+2x+2=0[/tex]
The solution to this equation can be found using the formula;
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where [tex]a=3,b=2,c=2[/tex]
We substitute these values to obtain;
[tex]x=\frac{-2\pm\sqrt{2^2-4(3)(2)}}{2(3)}[/tex]
This implies that;
[tex]x=\frac{-2\pm\sqrt{4-24}}{6}[/tex]
[tex]x=\frac{-2\pm\sqrt{-20}}{6}[/tex]
[tex]x=\frac{-2\pm2\sqrt{-5}}{6}[/tex]
[tex]x=\frac{-1\pm\sqrt{5}i}{3}[/tex]
Hence the solution is;
[tex]x=\frac{-1+\sqrt{5}i}{3}[/tex] or [tex]x=\frac{-\sqrt{5}i-1}{3}[/tex]
