Question

I really need help here ASAP​

Answer 1

D.
`x=(-4-√(31))/(3)` or
`x=(-4+√(31))/(3)`

Explanation:

The given equation is:

`3x^2+8x=5`

Divide through by 3;

`x^2+(8)/(3)x=(5)/(3)`

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

`x^2+(8)/(3)x+((4)/(3))^2=(5)/(3)++((4)/(3))^2`

`x^2+(8)/(3)x+(16)/(9)=(5)/(3)+(16)/(9)`

The left hand side is now a perfect square:

`(x+(4)/(3))^2=(31)/(9)`

Take square root

`x+(4)/(3)=\pm \sqrt{ (31)/(9)}`

`x=-(4)/(3)\pm \sqrt{ (31)/(9)}`

`x=-(4)/(3)\pm (√(31))/(3)`

D.
`x=(-4-√(31))/(3)` or
`x=(-4+√(31))/(3)`