Y=(X-3)(4x+1)

Solution for Y=(X-3)(4x+1) equation:

=(Y-3)(4Y+1)

We move all terms to the left:

-((Y-3)(4Y+1))=0

We multiply parentheses ..

-((+4Y^2+Y-12Y-3))=0


We calculate terms in parentheses: -((+4Y^2+Y-12Y-3)), so:

(+4Y^2+Y-12Y-3)

We get rid of parentheses

4Y^2+Y-12Y-3

We add all the numbers together, and all the variables

4Y^2-11Y-3

Back to the equation:

-(4Y^2-11Y-3)


We get rid of parentheses

-4Y^2+11Y+3=0

a = -4; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·(-4)·3
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*-4}=\frac{-24}{-8}
=+3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*-4}=\frac{2}{-8}
=-1/4
$