Y=(8x+3)(x-2)

Solution for Y=(8x+3)(x-2) equation:

=(8Y+3)(Y-2)

We move all terms to the left:

-((8Y+3)(Y-2))=0

We multiply parentheses ..

-((+8Y^2-16Y+3Y-6))=0


We calculate terms in parentheses: -((+8Y^2-16Y+3Y-6)), so:

(+8Y^2-16Y+3Y-6)

We get rid of parentheses

8Y^2-16Y+3Y-6

We add all the numbers together, and all the variables

8Y^2-13Y-6

Back to the equation:

-(8Y^2-13Y-6)


We get rid of parentheses

-8Y^2+13Y+6=0

a = -8; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·(-8)·6
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*-8}=\frac{-32}{-16}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*-8}=\frac{6}{-16}
=-3/8
$