Y=(3x+1)(x+3)

Solution for Y=(3x+1)(x+3) equation:

=(3Y+1)(Y+3)

We move all terms to the left:

-((3Y+1)(Y+3))=0

We multiply parentheses ..

-((+3Y^2+9Y+Y+3))=0


We calculate terms in parentheses: -((+3Y^2+9Y+Y+3)), so:

(+3Y^2+9Y+Y+3)

We get rid of parentheses

3Y^2+9Y+Y+3

We add all the numbers together, and all the variables

3Y^2+10Y+3

Back to the equation:

-(3Y^2+10Y+3)


We get rid of parentheses

-3Y^2-10Y-3=0

a = -3; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·(-3)·(-3)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8}{2*-3}=\frac{2}{-6}
=-1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8}{2*-3}=\frac{18}{-6}
=-3
$