9(1-3y)=4+y(3-y)

Solution for 9(1-3y)=4+y(3-y) equation:

9(1-3y)=4+y(3-y)

We move all terms to the left:

9(1-3y)-(4+y(3-y))=0

We add all the numbers together, and all the variables

9(-3y+1)-(4+y(-1y+3))=0

We multiply parentheses

-27y-(4+y(-1y+3))+9=0


We calculate terms in parentheses: -(4+y(-1y+3)), so:

4+y(-1y+3)

determiningTheFunctionDomain
y(-1y+3)+4

We multiply parentheses

-1y^2+3y+4

Back to the equation:

-(-1y^2+3y+4)


We get rid of parentheses

1y^2-3y-27y-4+9=0

We add all the numbers together, and all the variables

y^2-30y+5=0

a = 1; b = -30; c = +5;
Δ = b2-4ac
Δ = -302-4·1·5
Δ = 880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-4\sqrt{55}}{2*1}=\frac{30-4\sqrt{55}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+4\sqrt{55}}{2*1}=\frac{30+4\sqrt{55}}{2}
$