Question

How do you solve `x^2+3x-6=0`?

Answer 1

Find the zeros.

Explanation:

Solving implies we find the value(s) of `x`. This means we find the zeros of the function.

This equation is in standard form, so we have two methods to determining the value of `x`: complete the square or use the quadratic formula.

I will be using the quadratic formula because it's less confusing.

All we have to do is equate the equation to `0` (it's already done to us) and sub in the values accordingly. Then we solve.

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

`=(-[3]+- sqrt([3]^2 -4[1][-6]))/ (2[1])`

We simplify.

`=(-3+- sqrt(33))/ (2)`

Now we solve.

`x [+] = 1.372281323`

`~= 1.37`

`x [-] = -4.372281323`

`~= -4.37`

Therefore, the zeros are `1.37` and `-4.37`.

Here is a graph for reference.

graph{y=x^2 + 3x -6 [-20.27, 20.28, -10.14, 10.13]}

Hope this helps :)