How do you solve #12x^2-15x= 0# using the quadratic formula?

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Answer 1 · 2016-06-19T14:14:14

#x=0 and x=5/4#

Before you apply the quadratic formula, consider

a=12
b=-15
c=0

so, by substituting in the formula

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

you find

#x=(15+-sqrt(225-4*12*0))/(2*12)#

#x=(15+-15)/24#

#x=0 and x=30/24=5/4#