Y^3+y^2-25y=50

Solution for Y^3+y^2-25y=50 equation:

^3+Y^2-25Y=50

We move all terms to the left:

^3+Y^2-25Y-(50)=0

determiningTheFunctionDomain
Y^2-25Y-50+^3=0

We add all the numbers together, and all the variables

Y^2-25Y=0

a = 1; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·1·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*1}=\frac{0}{2}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*1}=\frac{50}{2}
=25
$