(Y^3+5y^2-2y+24)/y-2=0

Solution for (Y^3+5y^2-2y+24)/y-2=0 equation:

(^3+5Y^2-2Y+24)/Y-2=0


Domain of the equation: Y!=0

Y∈R


We multiply all the terms by the denominator


(^3+5Y^2-2Y+24)-2*Y=0

We add all the numbers together, and all the variables

(^3+5Y^2-2Y+24)-2Y=0

We get rid of parentheses

5Y^2-2Y-2Y+24+^3=0

We add all the numbers together, and all the variables

5Y^2-4Y=0

a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10}
=4/5
$