Question

Solve the equation 5x^{2}-3x-1= 0 to the nearest tenth.

Answer 1

`x=(√(29)+3 )/(10)` and
`( -√(29)+3 )/(10)`

Explanation:

Your equation is
`5x^(2)-3x-1=0`

But you need to put it in quadratic form, which is:

`x=\frac{-(b)±\\\sqrt{(b)^(2)-4(a)(c) }}{2(a)}` Ignore the Â, I couldn't get rid of it

If a=5, b=-3, and c=-1 then you just fill in the spaces

`x=\frac{-(-3)±\\\sqrt{(-3)^(2)-4(5)(-1) }}{2(5)}`

Then you simplify the equation

`x=(3±\\√(9-20(-1) ))/(10)`

`x=(3±\\√(9+20))/(10)`

`x=(3±\\√(29))/(10)`

Then you separate the equation

`x=(3+\\√(29))/(10)\\x=(3-\\√(29))/(10)`

Then you switch the numerators numbers

`x=(-\\√(29)+3)/(10)\\x=(\\√(29)+3)/(10)`

And that is your solution.

Answer 2

x=0.8 or x= -0.2

I hope this helps

:)