Answer:
5/12 and -11/12
Step-by-step explanation:
We need to solve the equation using quadratic formula but before that, there should be only one constant term so,
[tex]x^2+\frac{1x}{2} + \frac{1}{16}= \frac{4}{9}\\\\x^2+\frac{1x}{2} = \frac{4}{9} - \frac{1}{16}\\\\x^2+\frac{1x}{2} = \frac{4*16 - 1*9}{144}\\\\x^2+\frac{1x}{2} = \frac{55}{144}\\\\x^2+\frac{1x}{2} -\frac{55}{144}= 0[/tex]
Now using quadratic formula to find the value of x
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-\frac{1}{2}\pm\sqrt{(\frac{1}{2})^2-4(1)(\frac{-55}{144})}}{2(1)}\\\\x=\frac{-\frac{1}{2}\pm\sqrt{\frac{1}{4}+(\frac{55}{36})}}{2}\\\x=\frac{-\frac{1}{2}\pm\sqrt{\frac{9+55}{36}}}{2}\\\x=\frac{-\frac{1}{2}\pm\sqrt{\frac{64}{36}}}{2}\\\\x=\frac{-\frac{1}{2}\pm{\frac{8}{6}}}{2}\\\\so\,\,x=\frac{-\frac{1}{2}+\frac{8}{6}}{2}\,\, and\,\, x=\frac{-\frac{1}{2}-\frac{8}{6}}{2}\\ \\[/tex]
Solving these equations,
[tex]x= \frac{-3+8}{12} \,\, and \,\, x=\frac{-3-8}{12}\\\\x= \frac{5}{12} \,\, and \,\, x=\frac{-11}{12}[/tex]
So, x= 5/12 and x= -11/12
