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

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Answer 1 · 2017-07-03T05:58:54

Find the zeros.

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 :)

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