Question

How do you solve `2y ^ { 2} - 5x - 12= 0`?

Answer 1

`y=+-sqrt(5/2x+6)`

Explanation:

Taking it 1 step at a time so that you can see better what is going on

Using first principles (short cut method taken from this approach)

`color(purple)("Step 1")`

Add `color(red)(5x)` to both sides

`color(blue)(2y^2-5x-12=0" "->" "2y^2color(red)(+5x)-5x-12=0color(red)(+5x))`

`" "2y^2" "+0" "-12=5x`

`" "2y^2-12=5x`

...................................................................................................

`color(purple)("Step 2")`

Add `color(red)(12)` to both sides

`color(blue)(2y^2-12=5x" "->" "2y^2color(red)(+12)-12=5xcolor(red)(+12) )`

`" "2y^2" "+0" "=5x+12`

`" "2y^2=5x+12`

...................................................................................................

`color(purple)("Step 3")`

Divide both sides by 2 (same as `color(red)(xx1/2)`)

`color(blue)(2y^2=5x+12" "->" "2/(color(red)(2))y^2=5/(color(red)(2))x+12/(color(red)(2))`

But `2/2=1` giving:`" "y^2=5/2x+6`

...................................................................................................

`color(purple)("Step 4")`

Square root both sides:`" "sqrt(y^2)=sqrt(5/2x+6)`

`" "y=+-sqrt(5/2x+6)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(purple)(ul("Foot note"))`

Using an example `(-2)xx(-2)=+4`
`" "(+2)xx(+2) = +4`

As both `(-2)xx(-2)" and "(+2)xx(+2)` both equal 4 then

`sqrt(4) = "plus or minus "2 -> +-2`

Which is why we end up with `y=+-sqrt(5/2x+6)`