Question

How do you solve `x^2-3x-12=0`?

Answer 1

`x~~5.275`
`x~~-2.275`

Explanation:

There are many ways to solve a quadratic equation.
[Look here if you are lost on how to solve quadratic equations.](http://www.sosmath.com/algebra/solve/solve4/s43/s43.html)

When in doubt you can't go wrong with using the quadratic formula.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

where `ax^2+bx+c=0`

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For `x^2-3x-12=0`

`a=1`
`b=-3`
`c=-12`

Now just substitute into the quadratic formula

`x=(-(-3)+-sqrt((-3)^2-4(1)(-12)))/(2(1))`

`x=(3+-sqrt(9+48))/(2)`

`x=(3+-sqrt(57))/(2)`

Therefore,

`x=(3+sqrt(57))/(2)` `--->` `~~5.275`

OR

`x=(3-sqrt(57))/(2)` `--->` `~~-2.275`