How do you solve #y^2+3/2y=10#?

2 Answers
Jul 6, 2018

Answer:

#color(violet)(y = -4, 5/2#

Explanation:

#y^2 + (3/2) y = 10#

#2y^2 + 3x = 20#

#2y^2 + 3x - 20 = 0#

#2y^2 + 8y - 5y - 20 = 0#

#2y (y + 4) - 5(y +4) = 0#

#(y+4) * (2y-5) = 0#

#y = -4, 5/2#

Jul 6, 2018

Answer:

#y=-4" or "y=5/2#

Explanation:

#"subtract 10 from both sides"#

#y^2+3/2y-10=0#

#"multiply through by 2"#

#2y^2+3y-20=0#

#"factor the quadratic using the a-c method"#

#"the factors of the product "2xx-20=-40#

#"which sum to "+3" are "+8" and "-5#

#"split the middle term using these factors"#

#2y^2+8y-5y-20=0larrcolor(blue)"factor by grouping"#

#color(red)(2y)(y+4)color(red)(-5)(y+4)=0#

#"take out the "color(blue)"common factor "(y+4)#

#(y+4)(color(red)(2y-5))=0#

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

#y+4=0rArry=-4#

#2y-5=0rArry=5/2#