Question

How do you solve `3x^2+10x+2=0`?

Answer 1

`x = -5/3+-sqrt(19)/3`

Explanation:

One way involves completing the square. To avoid fractions a little, premultiply by `3` first, then use the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(3x+5)` and `b=sqrt(19)` as follows:

`0 = 3(3x^2+10x+2)`

`=9x^2+30x+6`

`=(3x)^2+2(3x)(5)+6`

`=(3x+5)^2-25+6`

`=(3x+5)^2-19`

`=(3x+5)-(sqrt(19))^2`

`=((3x+5)-sqrt(19))((3x+5)+sqrt(19))`

`=(3x+5-sqrt(19))(3x+5+sqrt(19))`

`=9(x+5/3-sqrt(19)/3)(x+5/3+sqrt(19)/3)`

Hence:

`x = -5/3+-sqrt(19)/3`