Y=(4x-2)(2x+1)

Solution for Y=(4x-2)(2x+1) equation:

=(4Y-2)(2Y+1)

We move all terms to the left:

-((4Y-2)(2Y+1))=0

We multiply parentheses ..

-((+8Y^2+4Y-4Y-2))=0


We calculate terms in parentheses: -((+8Y^2+4Y-4Y-2)), so:

(+8Y^2+4Y-4Y-2)

We get rid of parentheses

8Y^2+4Y-4Y-2

We add all the numbers together, and all the variables

8Y^2-2

Back to the equation:

-(8Y^2-2)


We get rid of parentheses

-8Y^2+2=0

a = -8; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-8)·2
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{-(0)-8}{2*-8}=\frac{-8}{-16}
=1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-8}=\frac{8}{-16}
=-1/2
$