How do you solve #-3x^2=2(3x-5)# using the quadratic formula?

1 Answer
Jan 22, 2017

Answer:

Rearrange to form: #3x^2+6x-10=0#
Substitute in quadratic formula: #x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))#
Simplify to get: #x=(-3+sqrt(39))/(3) ~~1.08167# or #x=(-3-sqrt(39))/(3)~~-3.08167#

Explanation:

First of all you need to rearrange the equation into the form #ax^2+bx+c=0#

So first expand the brackets on the right hand side: #-3x^2=6x-10#

Then you can either take the #-3x^2# over or the #6x-10# over.

Let's just take the #-3x^2# over to give: #3x^2+6x-10=0#

Now we can see what #a# #b# and #c# are:
#a=3#
#b=6#
#c=-10#

Then we substitute these into the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Quadratics usually have 2 answers which is why there is a #+-# as for one answer you use #+# and the other #-#.

So now substituting in: #x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))#

This simplifies to: #x=(-6+-2sqrt(39))/(6)#

And dividing all terms by 2 to: #x=(-3+-sqrt(39))/(3)# which is the most simple form.

So #x=(-3+sqrt(39))/(3) ~~1.08167# or #x=(-3-sqrt(39))/(3)~~-3.08167#