How do you solve #3x^2+27x=108#?

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Answer 1 · 2016-08-26T14:50:15

The solutions are the same as the solutions of #3x^2+27x-108=0#

You cam make this equation a bit simpler by dividing it by 3. The solutions don't change. So, the modified equation is:

#x^2+9x-36#

Using the formula for the solutions of #ax^2+bx+c=0#, as follows:

#x_(1,2)=(-b+-sqrt(b^2-(4*a*c)))/(2a)#

we have in this case:

#x_(1,2)=(-9+-sqrt(81-(4*1*(-36))))/2#, and then we have:

#x_(1,2)=(-9+-sqrt(81+144))/2=(-9+-sqrt(225))/2=(-9+-15)/2#

Thus, we have the solutions:

#x_1=(-9+15)/2=6/2=3#

#x_2=(-9-15)/2=-24/2=-12#