Answer:
D. [tex]x=\frac{-4-\sqrt{31}}{3}[/tex] or [tex]x=\frac{-4+\sqrt{31}}{3}[/tex]
Step-by-step explanation:
The given equation is:
[tex]3x^2+8x=5[/tex]
Divide through by 3;
[tex]x^2+\frac{8}{3}x=\frac{5}{3}[/tex]
Add the square of half the coefficient of x to both sides.
[tex]x^2+\frac{8}{3}x+(\frac{4}{3})^2=\frac{5}{3}++(\frac{4}{3})^2[/tex]
[tex]x^2+\frac{8}{3}x+\frac{16}{9}=\frac{5}{3}+\frac{16}{9}[/tex]
The left hand side is now a perfect square:
[tex](x+\frac{4}{3})^2=\frac{31}{9}[/tex]
Take square root
[tex]x+\frac{4}{3}=\pm \sqrt{ \frac{31}{9}}[/tex]
[tex]x=-\frac{4}{3}\pm \sqrt{ \frac{31}{9}}[/tex]
[tex]x=-\frac{4}{3}\pm \frac{\sqrt{31}}{3}[/tex]
D. [tex]x=\frac{-4-\sqrt{31}}{3}[/tex] or [tex]x=\frac{-4+\sqrt{31}}{3}[/tex]
