How do you solve `6x^2-21x+15=0`?

Start by simplifying your equation by factoring out a 3:
`3(2x^2-7x+5)=0`
`2x^2-7x+5=0`

This equation cannot be factored with whole numbers, so you should use the quadratic formula:

`(-b+-sqrt(b^2-4ac))/(2a)`, knowing that `ax^2+bx+c`

So now:
`(-(-7)+-sqrt((-7)^2-4(2)(5)))/(2(2))`
`(7+-sqrt(49-4(2)(5)))/(4)`
`(7+-sqrt(49-40))/(4)`
`(7+-sqrt(9))/(4)`
`(7+-3)/(4)`
`10/4` or `4/4`=
`5/2` or `1`

x= `5/2` or `1`

In order to complete the square move the last term (term without `x`) to other side of equation
`x^2-21/6x=-15/6`

Then you want to find a piece that allows you to find a square square of the left hand side
i.e. `a^2+2ab+b^2=(a+b)^2`
or
`a^2-2ab+b^2=(a-b)^2`

In this equation `x=a`, `2ab=-21/6x` so as `x=a` we know that `2b=-21/6` so to complete the square we just need `b^2` so if we half and square `2b` we will get it so `b^2=(21/12)^2`

So if we add this term to both sides we get

`x^2-21/6x+(21/12)^2=-15/6+(21/12)^2`
Now the left hand side can be simplifed into merely `(a-b)^2`

`(x-21/12)^2=-15/6+441/144`

`(x-21/12)^2=-15/6+49/16`

Find a common multiple for 16 and 6 and add them together

`(x-21/12)^2=-240/96+294/96`

`(x-21/12)^2=54/96`

Square root both sides

`x-21/12=+-sqrt(54/96)`

`x=21/12+-sqrt(54/96)`