Question

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

Answer 1

`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`

Answer 2

`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`