How do you solve #4x^2-6=-12x#?

1 Answer
Dec 27, 2016

Answer:

#x=(-3+-sqrt(15))/2#

Explanation:

#4x^2-6=12x#

#rarr 4x^2+12x-6=0#

#rarr 2x^2+6x-3=0#

applying the quadratic formula:
#color(white)("XXX")x=(-6+-sqrt(6^2-4 * 2 * (-3)))/(2 * 2)#

#color(white)("XXX")=(-6+-sqrt(36+24))/4#

#color(white)("XXX")=(-cancel(6)^3+-cancel(2)sqrt(15))/(cancel(4)_2)#

You could use a calculator if you needed this in an approximate decimal form:
#color(white)("XXX")(-3+-sqrt(15))/2~~0.43691673 or -3.436491673#