Question

I really need help here ASAP​

Answer 1

Answer:

D. $x=\frac{-4-\sqrt{31}}{3}$ or  $x=\frac{-4+\sqrt{31}}{3}$

Step-by-step explanation:

The given equation is:

$3x^2+8x=5$

Divide through by 3;

$x^2+\frac{8}{3}x=\frac{5}{3}$

Add the square of half the coefficient of x to both sides.

$x^2+\frac{8}{3}x+(\frac{4}{3})^2=\frac{5}{3}++(\frac{4}{3})^2$

$x^2+\frac{8}{3}x+\frac{16}{9}=\frac{5}{3}+\frac{16}{9}$

The left hand side is now a perfect square:

$(x+\frac{4}{3})^2=\frac{31}{9}$

Take square root

$x+\frac{4}{3}=\pm \sqrt{ \frac{31}{9}}$

$x=-\frac{4}{3}\pm \sqrt{ \frac{31}{9}}$

$x=-\frac{4}{3}\pm \frac{\sqrt{31}}{3}$

D. $x=\frac{-4-\sqrt{31}}{3}$ or  $x=\frac{-4+\sqrt{31}}{3}$