Question

How do you solve `5x^2+3x-4=0` by completing the square?

Answer 1

The roots are

`x_1=(-3+sqrt(89))/10`
`x_2=(-3-sqrt(89))/10`

Explanation:

From the given
`5x^2+3x-4=0`

factor out 5 from the first two terms

`5(x^2+3/5x)-4=0`

Use now the number 3/5. Divide this number by 2 then square the result to obtain 9/100. This 9/100 will be added and subtracted to terms inside the grouping symbol.

Let us continue

`5(x^2+3/5x)-4=0`

`5(x^2+3/5x+9/100-9/100)-4=0`

We now have a perfect square trinomial

`x^2+3/5x+9/100=(x+3/10)^2` so that

`5((x+3/10)^2-9/100)-4=0`

Transpose now the -4 to the right side then divide by 5 both sides of the equation then transpose the -9/100 to the right also.

`5((x+3/10)^2-9/100)=4`

`(cancel5((x+3/10)^2-9/100))/cancel5=4/5`

`(x+3/10)^2-9/100=4/5`

`(x+3/10)^2=4/5+9/100`

simplify

`(x+3/10)^2=(80+9)/100`

`(x+3/10)^2=89/100`
Extract the square root of both sides

`sqrt((x+3/10)^2)=+-sqrt(89/100)`

`x+3/10=+-sqrt(89/100)`

`x=-3/10+-sqrt(89)/10`

The roots are

`x_1=-3/10+sqrt(89)/10`
`x_2=-3/10-sqrt(89)/10`

God bless....I hope the explanation is useful.