3(y+3)+5y=y(2y+1)+5

Solution for 3(y+3)+5y=y(2y+1)+5 equation:

3(y+3)+5y=y(2y+1)+5

We move all terms to the left:

3(y+3)+5y-(y(2y+1)+5)=0

We add all the numbers together, and all the variables

5y+3(y+3)-(y(2y+1)+5)=0

We multiply parentheses

5y+3y-(y(2y+1)+5)+9=0


We calculate terms in parentheses: -(y(2y+1)+5), so:

y(2y+1)+5

We multiply parentheses

2y^2+y+5

Back to the equation:

-(2y^2+y+5)


We add all the numbers together, and all the variables

8y-(2y^2+y+5)+9=0

We get rid of parentheses

-2y^2+8y-y-5+9=0

We add all the numbers together, and all the variables

-2y^2+7y+4=0

a = -2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-2)·4
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*-2}=\frac{-16}{-4}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-2}=\frac{2}{-4}
=-1/2
$