What are the approximate solutions of #4x^2+3=-12x# to the nearest hundredth?

1 Answer
Meave60
Jun 2, 2017

#x=-0.28,-2.72#

Explanation:

#4x^2+3=-12x#

Move all terms to the left side.

#4x^2+3+12x=0#

Rearrange to standard form.

#4x^2+12x+3# is a quadratic equation in standard form: #ax^2+bx+c#, where #a=4#, #b=12#, and #c=3#. You can use the quadratic formula to solve for #x# (the solutions).

Since you want approximate solutions, we won't solve the quadratic formula all the way. Once your values are inserted into the formula, you can use your calculator to solve for #x#. Remember there will be two solutions.

Quadratic Formula

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

Insert the known values.

Since you want the approximate solutions for #x#, you can put this into your calculator to get the approximate solutions.

#x=((-12+sqrt((12^2)-4*4*3)))/(2*4)=-0.28#

#x=((-12-sqrt((12^2)-4*4*3)))/(2*4)=-2.72#

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