2y+(y-3)=18-2y(y+3)

Solution for 2y+(y-3)=18-2y(y+3) equation:

2y+(y-3)=18-2y(y+3)

We move all terms to the left:

2y+(y-3)-(18-2y(y+3))=0

We get rid of parentheses

2y+y-(18-2y(y+3))-3=0


We calculate terms in parentheses: -(18-2y(y+3)), so:

18-2y(y+3)

determiningTheFunctionDomain
-2y(y+3)+18

We multiply parentheses

-2y^2-6y+18

Back to the equation:

-(-2y^2-6y+18)


We add all the numbers together, and all the variables

-(-2y^2-6y+18)+3y-3=0

We get rid of parentheses

2y^2+6y+3y-18-3=0

We add all the numbers together, and all the variables

2y^2+9y-21=0

a = 2; b = 9; c = -21;
Δ = b2-4ac
Δ = 92-4·2·(-21)
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{249}}{2*2}=\frac{-9-\sqrt{249}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{249}}{2*2}=\frac{-9+\sqrt{249}}{4}
$