Question

Solve the equation:
(3y+2)(y+3)=y+14
Please show work

Answer 1

multiply out the left side then make 1 side zero. solve the quadratic.

Answer 2

Okay thanks

Answer 3

\[\Large (3y+2)(y+3)=y+14\]

multiply left side and you'll get...
\[\Large 3y^2+9y+2y+6=y+14\]

from here you go...

\[\Large 3y^2+11y+6-y-14=0\] we moved the left side to the right and changed signs...
now you have quadratic equation...

\[\Large 3y^2+10y-8=0\]


Now use this formula...
\[\LARGE x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

you have

a=3 b=10 and c=-8 just substitute and there you go !... ;)