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 :)