y=8y(4y+32)=16

Solution for y=8y(4y+32)=16 equation:

y=8y(4y+32)=16

We move all terms to the left:

y-(8y(4y+32))=0


We calculate terms in parentheses: -(8y(4y+32)), so:

8y(4y+32)

We multiply parentheses

32y^2+256y

Back to the equation:

-(32y^2+256y)


We get rid of parentheses

-32y^2+y-256y=0

We add all the numbers together, and all the variables

-32y^2-255y=0

a = -32; b = -255; c = 0;
Δ = b2-4ac
Δ = -2552-4·(-32)·0
Δ = 65025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{65025}=255$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-255)-255}{2*-32}=\frac{0}{-64}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-255)+255}{2*-32}=\frac{510}{-64}
=-7+31/32
$