Y=8x(4x+1)

Solution for Y=8x(4x+1) equation:

=8Y(4Y+1)

We move all terms to the left:

-(8Y(4Y+1))=0


We calculate terms in parentheses: -(8Y(4Y+1)), so:

8Y(4Y+1)

We multiply parentheses

32Y^2+8Y

Back to the equation:

-(32Y^2+8Y)


We get rid of parentheses

-32Y^2-8Y=0

a = -32; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-32)·0
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{-(-8)-8}{2*-32}=\frac{0}{-64}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-32}=\frac{16}{-64}
=-1/4
$