Question 192498
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Given: 3x(x+3)=2(5x+1)

Expand:
{{{3x^2+9x=10x+2}}}
{{{3x^2+9x-10x-2=0}}}
{{{3x^2-x-2=0}}}, where{{{system(a=3,b=(-1),c=(-2))}}}


Then,
{{{x=(-b+-sqrt(b^2-4ac))/(2a)=(-(-1)+-sqrt(-1^2-4(3)(-2)))/(2*3)}}}
{{{x=(1+-sqrt(1+24))/6=(1+-sqrt(25))/6=(1+-5)/6}}}


Two values:
{{{x=(1+5)/6=6/6=red(1)}}}
{{{x=(1-5)/6=-4/6=red(-2/3)}}}


Let us check: x = 1
{{{3x(x+3)=2(5x+1)}}}
{{{3(1)(1+3)=2(5*1+1)}}}
{{{3*4=2*6}}}
{{{12=12}}}, good


x = -2/3
{{{3(-2/3)(-2/3+3)=2(5(-2/3)+1)}}}
{{{(-2)((-2+9)/3)=2(-10/3+1)}}}
{{{(-2)(7/3)=2((-10+3))/3}}}
{{{-14/3=2(-7/3)}}}
{{{-14/3=-14/3}}}, good


Let us see the graph:
{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5,3x^2-x-2),blue(circle(-2/3,0,.10)),blue(circle(1,0,.10)),blue(circle(-2/3,0,.07)),blue(circle(1,0,.07)))}}}


Thank you,
Jojo</font>