Y^3-100y^2+2500y=128

Solution for Y^3-100y^2+2500y=128 equation:

^3-100Y^2+2500Y=128

We move all terms to the left:

^3-100Y^2+2500Y-(128)=0

determiningTheFunctionDomain
-100Y^2+2500Y-128+^3=0

We add all the numbers together, and all the variables

-100Y^2+2500Y=0

a = -100; b = 2500; c = 0;
Δ = b2-4ac
Δ = 25002-4·(-100)·0
Δ = 6250000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6250000}=2500$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2500)-2500}{2*-100}=\frac{-5000}{-200}
=+25
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2500)+2500}{2*-100}=\frac{0}{-200}
=0
$