How do you solve #5r+3=2r^2#?

2 Answers
Aug 8, 2018

Answer:

#r_1=-1/2# and #r_2=3#

Explanation:

#5r+3=2r^2#

#2r^2-5r-3=0#

#(2r+1)*(r-3)=0#

So #r_1=-1/2# and #r_2=3#

Aug 8, 2018

Answer:

#r=-1/2" or "r=3#

Explanation:

#"rearrange into standard form ";ax^2+bx+c=0;a!=0#

#"subtract "5r+3" from both sides"#

#2r^2-5r-3=0#

#"factor using the a-c method"#

#"the factors of the product "2xx-3=-6#

#"which sum to "-5" are "-6" and "+1#

#"split the middle term using these factors"#

#2r^2-6r+r-3=0larrcolor(blue)"factor by grouping"#

#color(red)(2r)(r-3)color(red)(+1)(r-3)=0#

#"take out the "color(blue)"common factor "(r-3)#

#(r-3)(color(red)(2r+1))=0#

#"equate each factor to zero and solve for r"#

#2r+1=0rArrr=-1/2#

#r-3=0rArrr=3#