How do you solve #3x^2+5x=2# by using the quadratic formula?

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Answer 1 · 2016-10-06T08:56:45

#x_1=-2#
#x_2=1/3#

Given a quadratic equation

#ax^2+bx+c=0#

the Quadratic Formula tell us that:

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

You need to lead your equation in the canonical form:

#3x^2+5x=2=>3x^2+5x-2=0#

Therefore you find:

#a=3; b=5; c=-2#

So you can substitute them in the quadratic formula:

#x_(1,2)=(-5+-sqrt(5^2-4*3*(-2)))/(2*3)=(-5+-sqrt(25+24))/6=#
#=(-5+-sqrt(49))/6=(-5+-7)/6#

#x_1=(-5-7)/6=-cancel(12)^2/cancel(6)_1=-2#
#x_2=(-5+7)/6=cancel(2)^1/cancel(6)_3=1/3#

graph{3x^2+5x-2 [-2.434, 2.433, -1.214, 1.218]}